Showing that $f(x)=x^2$ for $x \in \mathbb{Q}$ and $f(x)=0$ for $x \not\in \mathbb{Q}$ is differentiable in $x=0$ I am supposed to show that $f(x) = x^2$ for $x$ in the rationals and $f(x) = 0$ for $x$ in the irrationals is differentiable at $x = 0$ and I am supposed to find the derivative of $f(x)$ at $x = 0$.
Is my proof correct or not?
My proof:
consider limit as $h\rightarrow0$ of $\frac{f(0 + h) - f(0)}h$ 
then we have limit as $h\rightarrow0$ of $\frac{h^2 - 0^2}h$
and then we get limit as $h\rightarrow0$ of $\frac{h^2}h
=$ limit as $h\rightarrow0$ of $h= 0 = $ the derivative of $f(x)$ at $0$

You're pretty close, but you're missing the case when $x$ is irrational.
To find the derivative we have to evaluate $$f'(0)=\lim\limits_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=\lim\limits_{x\rightarrow 0}\frac{f(x)}{x}$$ since $f(0)=0^2=0$. One way to evaluate this is to let $(x_n)\rightarrow 0$ be an arbitrary sequence converging to zero with $x_n\neq 0$ for all $n$. Then if $$\lim\limits_{n\rightarrow\infty}\frac{f(x_n)-f(0)}{x_n-0}=\lim\limits_{n\rightarrow\infty}\frac{f(x_n)}{x_n}=0$$ we're done. Let $\varepsilon>0$ be given. We're trying to find $N\in \mathbb{N}$ such that $\forall\ n\geq N$, $\frac{f(x_n)}{x_n}<\varepsilon$. Note that since $(x_n)\rightarrow 0$, $\exists\ N\in\mathbb{N}$ such that $\forall\ n\geq N, |x_n|<\varepsilon$. Now let $n\geq N$. Let's evaluate $\frac{f(x_n)}{x_n}$.
Case 1 $x_n\in\mathbb{Q}$. Then $$\frac{f(x_n)}{x_n}=\frac{(x_n)^2}{x_n}=x_n<\varepsilon\  \checkmark$$
Case 2 $x_n\notin\mathbb{Q}$. Then 
$$\frac{f(x_n)}{x_n}=\frac{0}{x_n}=0<\varepsilon\ \checkmark$$
Thus $f$ is differentiable at $0$ and $$f'(0)=\lim\limits_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=0$$
