function positive, if positive on axis I'm stuck on this little problem, and I'd appreciate some help
shouldn't be too hard, but I can't really get my head around it:

Let $f: \mathbb R^2 \to \mathbb R$ be a $C^1$ function such that \begin{cases}
\frac{\partial f}{\partial t}(x,t) = \frac{\partial f}{\partial x}(x,t), & \forall \, (x,t) \in \mathbb R^2 \\
f(x,0) > 0, & \forall \, (x,0) \in \mathbb R^2
\end{cases}
Show that $f(x,t) >0$, $\forall \, (x,t) \in \mathbb R^2$.

By continuity I can show that there exists a region around the $x$-axis where this is true.
I'm not sure though how to extend this idea!
 A: This answer might be overkill. Feel free to share an easier solution, I am eager to see one.
The assumption on the partial derivatives implies the existence of $g\colon\mathbb{R}\rightarrow\mathbb{R}$ such that for all $(x,t)\in\mathbb{R}^2$:
$$f(x,t)=g(x+t).$$
The second assumption implies that $g$ is non-negative, whence the result.

Let us establish the result I used implicitely:

Proposition. Let $f\colon\mathbb{R}^m\rightarrow\mathbb{R}$ be a smooth function. Assume that all the partial derivatives of $f$ are equal, then there exists $g\colon\mathbb{R}\rightarrow\mathbb{R}$ such that for all $(x_1,\ldots,x_m)\in U$, one has:
  $$f(x_1,\ldots,x_m)=g(x_1+\cdots+x_m).$$

Proof. Let us prove that $f$ is constant on all the hyperplane of equations $x_1+\cdots+x_m=k$, where $k\in\mathbb{R}$. Doing so, we will have that:
$$f(x_1,\ldots,x_m)=f(x_1+\cdots+x_m,0,\ldots,0).$$
Therefore, $g(t)=f(t,0,\cdots,0)$ suits the requirement and the result follows.
Let $\mathbf{x}:=(x_1,\ldots,x_m)$ and $\mathbf{y}:=(y_1,\cdots,y_m)$ such that:
$$x_1+\cdots+x_m=y_1+\cdots+y_m.$$
Let define $x\colon [0,1]\rightarrow\mathbb{R}^m$ by:
$$x(t)=(1-t)\mathbf{x}+t\mathbf{y}.$$
Also, let $F:=f\circ x\colon [0,1]\rightarrow\mathbb{R}$, then $F$ is differentiable on $[0,1]$ and using chain rule, one has:
$$\dot{F}(t)=\sum_{k=1}^m(y_k-x_k)\frac{\partial f}{\partial x_k}(x(t))=\left(\sum_{k=1}^my_k-\sum_{k=1}^mx_k\right)\frac{\partial f}{\partial x_1}(x(t))=0.$$
Hence, $f(\mathbf{x})=F(0)=F(1)=f(\mathbf{y})$. $\Box$.
A: One of my students (I'm a TA) had the solution in like one minute after I posed the problem. It's the same idea as posted by C. Falcon above, just with a different path, more 'natural' to the problem.
the problem reduces to showing that $f(x_1,\cdots,x_n) = f(\sum_{i=1}^n x_i,0,\cdots,0)$
let $\gamma : [0,1] \to \mathbb R^n$ the path connecting $\mathbf x$ and $\mathbf y$ where
$$ \gamma (0) = \mathbf x = \begin{pmatrix} \sum_{i=1}^n x_i \\ 0 \\ \vdots \\ 0 \end{pmatrix}, 
\qquad \gamma(1) = \mathbf y = \begin{pmatrix}  x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}, \\
\qquad \gamma(s) = \begin{pmatrix}  \sum_{i=1}^n x_i - s \, \sum_{i=1}^{n-1} x_i \\ s \, x_2 \\ \vdots \\ s \, x_n \end{pmatrix}, 
\qquad \dot \gamma = \begin{pmatrix} - \sum_{i=1}^{n-1} x_i \\ x_2 \\ \vdots \\ x_n \end{pmatrix}$$ 
then $$f(\gamma(0))- f(\gamma(1)) = \int_0^1 \nabla f \cdot \dot \gamma \, dt = 0 $$
since $\nabla f \propto (1,1,\cdots,1)^T$ and $\dot \gamma \perp (1,1,\cdots,1)^T$
