Let $f,g\colon\mathbb{R}\to\mathbb{R}$ be differentiable functions. Is it true that the function $h\colon\mathbb{R}\to\mathbb{R}$ defined by
$$ h(x) = \begin{cases} \dfrac{f(x)g(x)^2}{f(x)^2+g(x)^2} & \text{if } (f(x),g(x)) \neq (0,0) \\ 0 & \text{if } (f(x),g(x)) = (0,0). \end{cases} $$
is differentiable everywhere? It is clear that $h$ is differentiable whenever $(f(x),g(x))\neq(0,0)$, by the sum, product and quotient rules. The derivative however is quite messy and it doesn't seem clear how to proceed efficiently to deal with the origin.
Also the function from $\mathbb{R}^2\to\mathbb{R}$ $$(x,y)\mapsto \frac{xy^2}{x^2+y^2}$$ is not differentiable at the origin, so an approach which uses the chain rule seems unlikely to work. An elementary solution not invoking heftier theorems, such as L'Hospitals's rule, would be preferred.
Assume that $f(0)=g(0)=0$. Then $$h(x)=f(x)\>{g^2(x)\over f^2(x)+g^2(x)}$$ is continuous at $0$, and $h(0)=0$. As $f$, $g$ are differentiable there are continuous functions $\phi$ and $\psi$ such that $$f(x)=x\phi(x),\quad \phi(0)=f'(0),\qquad g(x)=x\psi(x),\quad \psi(0)=g'(0)\ .$$ It follows that $${f(x)g^2(x)\over f^2(x)+g^2(x)}=x\>\phi(x)\>{\psi^2(x)\over\phi^2(x)+\psi^2(x)}=:x\> \Phi(x)\ .$$ If $(\phi(0),\psi(0))\ne(0,0)$ the limit $\lim_{x\to0}\Phi(x)$ obviously exists. If $\phi(0)=\psi(0)=0$ then $\lim_{x\to0}\Phi(x)=0$ since the large fraction is $\leq1$ and $\lim_{x\to0}\phi(x)=0$.
It follows that your function $h$ is not only continuous but even differentiable at $0$.