Show function $g$ is continuously differentiable 
Let $f\in C^2(\mathbb{R^2})$ with $f(0, λ) = 0$ $\forall \lambda \in \mathbb{R}$.
Show $$g(x,\lambda):=\begin{cases} \frac{f(x,\lambda)}{x}& \text{ if } x\ne0 \\ \partial_x f(0,\lambda)&\text{ if } x=0 \end{cases}$$ is continuously differentiable.

So I need to show that $g$ is continuous and its derivative is also a continuous function. Since $f\in C^2(\mathbb{R^2})$ and $\frac{1}{x}$ is continous on $\mathbb{R}\backslash\{0\}$, $g(x,\lambda)$ is continous as a composition/product of continuous functions in both cases and I only need to analyse the point $x=0$.
Since $f(0, λ) = 0$ I can rewrite the first case to $\frac{f(x,\lambda) - f(0,\lambda)}{x-0}$ which is the difference quotient in $0$ and one gets $$lim_{x\to0}\frac{f(x,\lambda) - f(0,\lambda)}{x-0} = \partial_xf(0,\lambda),\text{ so } g \text{ is continuous}.$$
Problem: I also need to show the continuity of its derivative: $$\partial_xg(x,\lambda)=\begin{cases} \frac{\partial_xf(x,\lambda)}{x}-\frac{f(x,\lambda)}{x^2}& \text{ if } x\ne0 \\ \partial_{xx} f(0,\lambda)&\text{ if } x=0 \end{cases} \text{ and } \partial_{\lambda}g(x,\lambda)=\begin{cases} \frac{\partial_{\lambda}f(x,\lambda)}{x}& \text{ if } x\ne0 \\ \partial_{\lambda}(\partial_x f(0,\lambda))&\text{ if } x=0 \end{cases}$$
$\partial_{\lambda}g(x,\lambda)$ should be the same argument as the continuity for $g$ just with a $\partial_{\lambda}$ in front of it since the cases of $g$ only depend on $x$.
Since $\partial_xf(0,\lambda)=lim_{x\to 0}\frac{f(x,\lambda)-f(0,\lambda)}{x}$ and $f(0,\lambda)=0$ $\forall \lambda$, it is $$\partial_{xx}f(0,\lambda)=lim_{x\to 0}\frac{\partial_xf(x,\lambda)-\partial_xf(0,\lambda)}{x}=lim_{x\to 0}\frac{\partial_xf(x,\lambda)}{x}-\frac{f(x,\lambda)}{x^2},$$ so the derivative of $g$ is also continuous.
Is my proof correct? Thanks!
 A: In writing out the cases for $\partial_{x}g(0,\lambda)$, you seem to have derived $$\partial_{x}g(0,\lambda)=\partial_{xx}f(0,\lambda)$$ from $g(0,\lambda)=\partial_{x}f(0,\lambda)$. But that is not the reason the equality above holds.
Let me show a counterexample where that reasoning fails. Define a function $$h(x):=\begin{cases}
0 & x\neq0\\
f'(0) & x=0
\end{cases}$$ and take $f(x)=\cos x$ (it took me a while to come up with this example! :D).
Note that $h(0)=\cos'(0)=-\sin(0)=0$ so that $h \equiv 0$. Hence, both $h$ and $f$ are even $C^\infty$, i.e. infinitely differentiable functions.
But it is not true that $h'(0)=f''(0)$ since $h'(0)=0$ but $f''(0)=-\cos(0)=-1$!
In your proof, you need to use the way $g(x)$ is defined for $x\neq 0$. It is important as I hope I made it clear with my counterexample.

Now about the specific problem in question. To show continuity of the derivative $\partial_{x}$, you need to compute $$\partial_{x}g(0,\lambda):=\lim_{x\to0}\frac{g(x,\lambda)-g(0,\lambda)}{x}$$ explicitly and show that it equals the limit $\lim_{x\to0}\left(\partial_{x}g(x,0)\right)$, and from your calculations you know $$\lim_{x\to0}\left(\partial_{x}g(x,0)\right)=\lim_{x\to0}\left(\frac{\partial_{x}f(x,\lambda)}{x}-\frac{f(x,\lambda)}{x^{2}}\right).$$
To finish, you need to show $\partial_{x}g(0,\lambda)=\lim_{x\to0}\partial_{x}g(x,\lambda)$, i.e. the following equality: $$\lim_{x\to0}\frac{g(x,\lambda)-g(0,\lambda)}{x}=\lim_{x\to0}\left(\frac{\partial_{x}f(x,\lambda)}{x}-\frac{f(x,\lambda)}{x^{2}}\right).$$
Which is trivial after you replace $g$ on the left with its values by definition.
