This is a homework problem, by the way.
Suppose $f: \mathbb{R} \rightarrow \mathbb{R} $ is differentiable at $x = 0$. Prove that $$\lim\limits_{x \rightarrow 0} \frac{f(x^2) - f(0)}{x} = 0$$
To say that $f$ is differentiable at $x = 0$ means that by the limit definition of the derivative, the limit $\lim\limits_{x \rightarrow 0} \frac{f(x) - f(0)}{x - 0}$ exists. However, I'm not sure where to proceed from here, or how that might help me prove the proposition.
On
Note $f(x^2)=(f\circ g)(x)$ with $g(x):=x^2$ and $g(x)$ is a differentiable function. Moreover $g(0)=0$ and $(f\circ g)(x)$ is differentiable at $0$. Let $h(x):=f(x^2)$ then $$\lim_{x\to 0}\frac{f(x^2)-f(0)}{x}=\lim_{x\to 0}\frac{h(x)-h(0)}{x}=h'(0)$$ But $h'(0)=f'(g(x))\cdot g'(x)\Big|_{x=0}=f'(x^2)\cdot2x\Big|_{x=0}=f'(0)\cdot 0=0$.
\begin{align*} \lim_{x\rightarrow 0}\dfrac{f(x^{2})-f(0)}{x}&=\lim_{x\rightarrow 0}\dfrac{f(x^{2})-f(0)}{x^{2}}\cdot x\\ &=\lim_{x\rightarrow 0}\dfrac{f(x^{2})-f(0)}{x^{2}}\cdot\lim_{x\rightarrow 0}x\\ &=f'(0)\cdot 0\\ &=0. \end{align*}