Is the composition of the two following functions differentiable. I was trying to prove it via the general defintion of differentiablity.
My result was 

$(h_{1}^{3}+h_{2}^{3})/(h_{1}^{2}+h_{2}^{2})^{(3/2)}$. However, i do not how to evaluate it beyond that point.

Moreover, is the function $f$ differentiable at $(x,y) = (0,0)$, because it seems like it isn't. I was thinking about proving this using partial derivatives, would this be the right way to go about it?
So the question has a typo: read    $\gamma_{a,b}(t)=(at,bt)$. If both $a,b$ are zero, this is a constant function, so the composition is also constant, hence differentiable.  Assume that not both $a,b$ are zero. Then for $t\ne 0$ we have 
$$f(\gamma_{a,b}(t))= \frac{(at)^3+(bt)^3}{(at)^2+(bt)^2} = t \frac{a^3+b^3}{a^2+b^2} \tag{1}$$ 
and for $t=0$, of course, $f(\gamma_{a,b}(t))=f(0,0)=0$. Thus, (1) holds for all $t$.
This is a linear function of $t$, hence differentiable.
So the question has a typo: read    $\gamma_{a,b}(t)=(at,bt)$. If both $a,b$ are zero, this is a constant function, so the composition is also constant, hence differentiable.  Assume that not both $a,b$ are zero. Then for $t\ne 0$ we have 
$$f(\gamma_{a,b}(t))= \frac{(at)^3+(bt)^3}{(at)^2+(bt)^2} = t \frac{a^3+b^3}{a^2+b^2} \tag{1}$$ 
and for $t=0$, of course, $f(\gamma_{a,b}(t))=f(0,0)=0$. Thus, (1) holds for all $t$.
This is a linear function of $t$, hence differentiable.