Show there exists $g:(0,1)\to\mathbb{R}^2$ such that $h\circ g$ constant, $g$ continuously differentiable and $g$ injective 
Let $h\in C^1(\mathbb{R}^2,\mathbb{R})$. Show:
  There exists $g:(0,1)\to\mathbb{R}^2$ such that $h\circ g$ constant, $g$ continuously differentiable and $g$ injective.

Dini's theorem:
Let $\Omega \subseteq \mathbb{R}^2$ open, $f \in C^1(\Omega, \mathbb{R})$ and $p \in \Omega$ with $f(p) = 0$.
If $\frac{\partial f}{\partial y}(p) \neq 0$, then there exist open intervals $V_1, V_2$ with $p \in V_1 \times V_2 \in \Omega$ and $g \in C^1(V, \mathbb{R})$ with
$$\{(x,y) \in V_1 \times V_2 \mid f(x,y) = 0 \} = \{(x,g(x) \mid x \in V_1\}$$
and $g$ furthermore satisfies
$$g' = -\left(\frac{\partial f}{\partial y}(x, g(x))\right)^{-1}\frac{\partial f}{\partial x}(x, g(x)).$$
Proof of existance of $g$:
Let without loss of generality $\frac{\partial f}{\partial y}(p) \gt 0$.
Then there exists $\epsilon \gt 0$ with $\overline{B_\epsilon(p)} \subseteq \Omega$ and $\frac{\partial f}{\partial y} \gt 0$ on $\overline{B_\epsilon(p)}$.
Let $y^- = p - \frac{\epsilon}{2}$ and $y^+ = p + \frac{\epsilon}{2}$. Then there exists $\delta \in (0, \epsilon)$ with $f(x, y^-) \lt 0 \lt f(x, y^+)$ for $x \in (p - \frac{\delta}{2}, p + \frac{\delta}{2}) =: V_1$ and $(y^-, y^+) =: V_2$.
For $x \in V_1$ there exists exactly one $y \in V_2$ with $f(x,y) = 0$ (mean value theorem). Set $y(x) := g$ and the claim follows.
 A: First of all, if $h$ is constant, you can take $g$ to be any smooth injective curve in $R^2$, e.g., $g(x)=(x,0)$. Therefore, in what follows I assume that $h$ is nonconstant. 
The way to solve the problem is to use the implicit function theorem; however, one has to be careful with the regularity assumptions: For a $C^1$-smooth function $h: R^2\to R$ and a point $p\in R^2$ such that $\nabla h(p)\ne 0$, $q=h(p)$, the implicit function theorem states that (after swapping the $x$ and $y$ coordinates in $R^2$ if necessary) there exists a $C^1$-function $f: R\to R$ and a neighborhood $U$ of $p$, such that the set $V=U\cap h^{-1}(q)$ is the graph of the function $f$ (restricted to some open interval $I\subset R$). The graph of $f|I$ admits the $C^1$-parameterization
$$
g(x)=(x, f(x)), x\in I. 
$$ 
Clearly, $g: I\to V$ is injective and $C^1$-smooth. It is also clear that $h\circ g$ is constant, equal $q\in R$. It remains, therefore, to find a point $p\in R^2$ such that $\nabla h(p)\ne 0$. By the mean value theorem, if $\nabla h=0$ everywhere on $R^2$, then $h$ is constant (the mean value theorem requires only $h$ to be differentiable). Thus, the point $p$ with $\nabla h(p)\ne 0$ exists unless $h$ is constant, which is already ruled out. 
A side remark. Whitney (see here) constructed an example of a $C^1$-function $h: R^2\to R$ such that there exists a (non-rectifiable) arc $\alpha$ in $R^2$ such that:


*

*$h|\alpha$ is nonconstant. 

*$\nabla h$ vanishes everywhere on $\alpha$.  
Therefore, in the above argument, one cannot take $q$ to be a "generic" real number. However, if we assume in addition that $h\in C^2(R^2,R)$ then Sard's theorem applies and one can take $q$ to be generic (in the measure-sense) and $p\in h^{-1}(q)$ to be arbitrary. 
A: Assume $h$ is not constant. Suppose $dh$ does not vanish at $x\in \mathbb{R}^2$. Try to use the implicit function theorem.
A: Here is a hopefully complete answer. Basically, this is just an elementary proof of the implicit function theorem for functions of 2 variables.
If $h$ is constant, you can take as $g$ any injective $\mathcal C^1$ map from $(0,1)$ into $\mathbb R^2$; for example $g(t)=(t,0)$.
Now, assume that $h$ is not constant. Then the partial derivatives of $h$ cannot be both identically $0$. So we may assume for example that $\frac{\partial h}{\partial y}(x_0,y_0)\neq 0$ for some point $(x_0,y_0)\in\mathbb R^2$; and without loss of generality, we may also assume that $\frac{\partial h}{\partial y}(x_0,y_0)> 0$. Set $$c:= f(x_0,y_0)\, .$$
Since $\frac{\partial h}{\partial y}$ is continuous, one can find an open interval $I_0\ni x_0$ and $\delta >0$ such that $$\frac{\partial h}{\partial y}(x,y)>0\quad {\rm for\; any\;}\; (x,y)\in I_0\times [y_0-\delta,y_0+\delta]\, .$$ 
The proof will be divided into several steps.
Step 1. One can find an open interval $I$ with $x_0\in I\subset I_0$ such that  and the following holds: for every $x\in I$, there is a unique point $y=y(x)\in [y_0-\delta,y_0+\delta]$ such that $h(x,y(x))=c$.
To prove this, observe first that the map $y\mapsto f(x_0,y)$ is increasing on $[y_0-\delta,y_0+\delta]$ because $\frac{\partial h}{\partial y}(x_0,y)>0$ on this interval by the choice of $\delta$. Since $f(x_0,y_0)=c$, it follows in particular that $h(x_0, y_0-\delta)<c<h(x_0,y_0+\delta)$. Now the set $$U=\{ x\in\mathbb R;\; h(x,y_0-\delta)<c<h(x,y_0+\delta)\} $$
is an open set in $\mathbb R$ by the continuity of $h$, and $x_0\in U$ by what has just been observed. So one can find an open interval $I_1$ such that $x_0\in I_1\subset U$. If we set $I:=I_0\cap I_1$ then $x_0\in I\subset I_0$ and $$h(x,y_0-\delta)<c<h(x,y_0+\delta)\quad{\rm for\; every\;}x\in I\, . $$
Let us fix any $x\in I$. Then the map $y\mapsto h(x,y)$ is continuous, and it is increasing on $[y_0-\delta, y_0+\delta]$ because $\frac{\partial h}{\partial y}(x,y)>0$ on this interval, and $h(x,y_0-\delta)<c<h(x,y_0+\delta)$. By the intyermediate value theorem, it follows that there exists a unique $y=y(x)\in [y_0-\delta, y_0+\delta]$ such that $h(x,y(x))=c$. This concludes Step 1.
Step 2. The map $x\mapsto y(x)$ is continuous on $I$.
We prove this by contradiction. Assume that this map is not continuous at some point $x\in I$. Then one can find a sequence $(x_n)\subset I$ converging to $x$ and $\varepsilon >0$ such that $\vert y(x_n)-y(x)\vert\geq \varepsilon$ for all $n\in\mathbb N$. Since $y(x_n)\in y_0-\delta,y_0+\delta]$, One can find a subsequence $(y(x_{n_k}))$ and a point $y\in [y_0-\delta,y_0+\delta]$ such that $y(x_{n_k})\to y$ (by Bolzano-Weierstrass). Bu we have $h(x_{n_k},y(x_{n_k}))=c$ for all $k$, so $h(x,y)=c$ because $h$ is continuous and $(x_{n_k},y(x_{n_k}))\to (x,y)$. Since $y\in [x_0-\delta,x_0+\delta]$, it follows that $y=y(x)$; but this is a contradiction since $\vert y(x_{n_k})-y(x)\vert\geq\varepsilon$ for all $k$ and hence $\vert y-y(x)\vert\geq\varepsilon >0$. This concludes Step 2.
Step 3. The map $x\mapsto y(x)$ is $\mathcal C^1$ on $I$.
It is enough to show that this map is differentiable at any point $x\in I$, with $$ y'(x)=-\frac{\frac{\partial h}{\partial x}(x,y(x))}{\frac{\partial h}{\partial y}(x,y(x))}\cdot$$
Indeed, since the partial derivatives of $h$ are continuous and the map $x\mapsto y(x)$ is continuous by Step 2, this formula will then show that $y'$ is continuous, i.e. that $y$ is $\mathcal C^1$.
Let us fix $x\in I$. Set $$a:=\frac{\partial F}{\partial x}(x,y(x))\quad{\rm and}\quad b:=\frac{\partial F}{\partial y}(x,y(x))$$
Keep in mind that $b\neq 0$ because $\frac{\partial F}{\partial y}(x,y)>0$ on $I\times [y_0-\delta,y_0+\delta]$.
By step 2, one may write
$$y(x+h)=y(x)+\varepsilon(h)\, ,$$ 
where $\varepsilon(h)\to 0$ as $h\to 0$. Moreover, we also have (by the differentiability of $h$ at $(x,y(x))$)
\begin{eqnarray}
h(x+h,y(x+h))&=&h(x+h, y(x)+\varepsilon (h))\\
&=&h(x,y(x))+ a\, h+b\, \varepsilon (h)+R(h)\, ,
\end{eqnarray}
where $R(h)=o(\vert h\vert+ \vert\varepsilon (h)\vert)$ as $h\to 0$, i.e.
$$\lim_{h\to 0}\frac{R(h)}{\vert h\vert+ \vert\varepsilon (h)\vert}=0\, . $$ 
Since $h(x+h,y(x+h))=c=h(x,y(x))$, this can be rewritten as
\begin{equation}\varepsilon(h)=-\frac{a}{b}\, h-\frac1b R(h).\end{equation}
If $h$ is small enough, we have $\left\vert-\frac1b R(h)\right\vert\leq \frac12\vert h\vert$ because $R(h)=o(\vert h\vert+\vert\varepsilon (h)\vert)$, so that $\vert\varepsilon(h)\vert\leq \left\vert\frac{a}{b}\right\vert \vert h\vert+\frac12(\vert h\vert+ \vert\varepsilon (h)\vert)$ and hence $\vert\varepsilon (h)\vert\leq C\,\vert h\vert$, where $C=2\vert a/b\vert+1$. This gives $\vert\varepsilon (h)\vert+\vert h\vert\leq (C+1)\vert h\vert$, which implies that in fact $$\lim_{h\to 0}\frac{R(h)}{\vert h\vert}=0\, . $$
Since \begin{eqnarray}y(x+h)&=&y(x)+\varepsilon(h)\\
&=&y(x)-\frac{a}{b}\, h-\frac1b R(h)\, ,\end{eqnarray}
we conclude that the map $y$ is indeed differentiable at $x$ with $y'(x)=-\frac{a}{b}\cdot$ This concludes Step 3.
Step 4. There is a one-to-one $\mathcal C^1$ map $g_0:I\to\mathbb R^2$ such that $h(g_0(x))=c$ for all $x\in I$.
Just define $g_0(x)=(x,y(x))$. This map is $\mathcal C^1$ by Step 3, it is clearly one-to-one because of its first coordinate, and $h(g_0(x))=c$ for all $x\in I$ by the very definition of $y(x)$.
Step 5. Conclusion.
Write $I=(\alpha,\beta)$ and define $g:(0,1)\to\mathbb R^2$ by $g(t)=g_0(\alpha+ t(\beta-\alpha))$. Then $g$ has the required properties.
A: This is not a full answer but it may provide some visual intuition for the problem.
If you visualize the graph of $h: \mathbb{R}^2 \rightarrow \mathbb{R}$ as a surface in the usual way, and imagine a plane of the form $z=c$ intersecting that surface, then what you want to do is choose a path $g:(0,1) \rightarrow \mathbb{R}^2$ that traces along part of the level curves where the plane cuts the surface.
