Differentiability and the chain rule (multivariable calculus)

I have a problem with the next exercise:

Given de function $\displaystyle f(x,y)=\begin{Bmatrix} \displaystyle\frac{xy^2}{x^2+y^2} & \mbox{ if }& (x,y)\neq{(0,0)}\\0 & \mbox{if}& (x,y)=(0,0)\end{matrix}$ with $\displaystyle \vec{g}(t)=\begin{Bmatrix} x=at \\y=bt \end{matrix},t\in{\mathbb{R}}$

a) Find $\displaystyle h=fog$ y $\displaystyle \displaystyle\frac{dh}{dt}$ for t=0

The thing is that I've found that f isn't differentiable at (0,0). The partial derivatives exists at that point, I've found them by definition.

$\displaystyle f_x(0,0)=0=f_y(0,0)$

And then I saw if it was differentiable at that point.

$\displaystyle \displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{xy^2}{(x^2+y^2)^{3/2}}}$

In the polar form it gives that this limit doesn't exists, so it isn't differentiable at that point. So I can't apply the chain rule there, right?

To ensure the differentiability of a composed function, both function must be differentiable. If one isn't, then the composition isn't differentiable at a certain point. Right?

Bye there, and thanks.