Tricky question about differentiability at the origin Let $f: \mathbb{R}^2 \to \mathbb{R} $ be given as 
$$
f(x,y) =
\begin{cases}
y, & \text{if }\text{ $x^2 = y $} \\
0, & \text{if }\text{ $x^2 \neq y $}
\end{cases}
$$
Is this function differentiable at $(0,0)$ ?

Yeah. It is easy to show that $f$ admits partial derivatives at $(0,0)$, both equal to $0$, for example $$\frac{\partial f}{\partial x}(0,0)=\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x}=0$$ 
To show that $\lim_{(x,y)\to 0}\frac{f(x,y)}{||(x,y)||}=0$ note that $$\Bigg|\frac{f(x,y)}{||(x,y)||}\Bigg|\leq \frac{x^2}{||(x,y)||}.$$ 
since $f(x,y)=x^2$ or$f(x,y)=0$. Now try to show that $$\lim_{(x,y)\to(0,0)}\frac{x^2}{\sqrt{x^2+y^2}}=0.$$ See the graph (in red) below to understand the definition of $f$

