Proving the Implicit function theorem in a particular case. 
Let $g:\Bbb{R^2}\rightarrow\Bbb{R}$ differentiable. Assume that $g(0,0)=0$ and $g'_y$ continuous at $(0,0)$ and $g'_y(0,0)>0$.
It is asking to prove the Implicit function theorem in this case.

My try :
By continuity, we have that $g'_y(x,y)>0$ on $[-\gamma,\gamma]$ for $\gamma>0$, as $g(0,0)=0$ we have $g(0,\gamma)>0$ and $g(0,-\gamma)<0$. Now by continuity of $g$, we can choose $\varepsilon>0$ and $<\gamma$ such that for all $x\in (-\varepsilon,\varepsilon):\quad g(x,\gamma)>0$ and $g(x,-\gamma)<0$.Then by the IVT for all $x\in (-\varepsilon,\varepsilon)$ there exist a unique $y\in (-\gamma,\gamma)$ such that $g(x,y)=0$.
Therefore, we have the existence of a function $\psi:(-\varepsilon,\varepsilon)\times [-\gamma,\gamma]$ such that $g(x,y)=0\Leftrightarrow y=\psi(x)$.
I need to prove that $\psi$ is differentiable, but here I am stuck.
 A: $\psi$ is continuous by construction. $\forall \gamma>0$ you can find $\epsilon$ small enough such that: $\psi((-\epsilon,\epsilon)) \subset (-\gamma,\gamma)$.
Suppose $g$ is $C^{1}$ on an open subset $U \in \mathbb{R}^{2}$. Assume $(-\epsilon, \epsilon) \times [-\gamma, \gamma] \subset U$ (ohterwise take $\gamma$ and $\epsilon$ smaller). The assumption that $g$ is $C^{1}$ on $U$ is needed.
Proof: Take $x,y \in (-\epsilon, \epsilon)$. Then: $$g(x,\psi(x)) =0 \text{ and } g(y,\psi(y)) =0$$
Consider: $$f:[0,1] \rightarrow \mathbb{R}: t \rightarrow g(x+t(y-x), \psi(x)+t \left[ \psi(y)-\psi(x) \right] )$$
First: $f$ is differentiable on $(0,1)$ ($g$ is differentiable on $U$) 
Seccond: $f(0)=0$ and $f(1) = 0$. Apply the mean value theorem on $f$. To get an element $T: 0<T<1$ and $ f'(T)=0$. Let $u=x+T.(y-x)$ and $v = \psi(x)+T.(\psi(y)-\psi(x))$.
Note: $$f'(t) = \frac{\partial g}{\partial x}(y-x)+\frac{\partial g}{\partial y}(\psi(y)-\psi(x))$$
So:
$$0=f'(T) = {\left( \frac{\partial g}{\partial x} \right)}_{(u,v)} (y-x)+{ \left( \frac{\partial g}{\partial y}  \right)}_{(u,v)} (\psi(y)-\psi(x))$$
We get:
$$-\frac{{\left( \frac{\partial g}{\partial x} \right)}_{(u,v)}}{{ \left( \frac{\partial g}{\partial y}  \right)}_{(u,v)}}= \frac{\psi(y)-\psi(x)}{y-x}$$
Note: ${ \left( \frac{\partial g}{\partial y}  \right)}_{(u,v)} \neq 0$.
Notice we can find a $(u,v)$ for every $y$. We denote this by: $(u_y,v_y)$
Next: Define:
$$H:(x-\delta,x+\delta) \rightarrow: \mathbb{R}: y \rightarrow \left\{ \begin{array}{lc}
             -\frac{{\left( \frac{\partial g}{\partial x} \right)}_{(x,\psi(x))}}{{ \left( \frac{\partial g}{\partial y}  \right)}_{(x,\psi(x))}}  & y=x \\
             \\ -\frac{{\left( \frac{\partial g}{\partial x} \right)}_{(u_{y},v_{y})}}{{ \left( \frac{\partial g}{\partial y}  \right)}_{(u_{y},v_{y})}} & y \neq 0 \\
             \end{array}
   \right.$$
Where we have taken a $\delta>0$ such that $(x-\delta,x+\delta) \subset(-\epsilon,\epsilon)$.
We have (see below): $$\mathop{lim}_{y \rightarrow x} H(y) = H(x)$$
Wich implies that $\psi$ is differentiable in $x$. This is true for all $x$ in $(-\epsilon,\epsilon)$.

(see below) explained:
$y \rightarrow x$ means: $(u_y,v_y) \rightarrow (x,\psi(x)) \in U$. Because $g$ is $C^{1}$ on $U$. It follows that 
$$-\frac{{\left( \frac{\partial g}{\partial x} \right)}_{(u_{y},v_{y})}}{{ \left( \frac{\partial g}{\partial y}  \right)}_{(u_{y},v_{y})}} \mathop{\rightarrow}^{y \rightarrow x} -\frac{{\left( \frac{\partial g}{\partial x} \right)}_{(x,\psi(x))}}{{ \left( \frac{\partial g}{\partial y}  \right)}_{(x,\psi(x))}} $$
