Differentiation of Vector Functions: Proving Differentiability with Chain Rule

[boombaby]

Homework Statement

Define f(0,0)=0 and $$f(x,y)=\frac{x^{3}}{x^{2}+y^{2}}$$ if (x,y)!=(0,0)
Let $$\gamma$$ be a differentiable mapping of R¹ into R², with $$\gamma(0)=(0,0)\;and\; |\gamma'(0)>0|$$. Put $$g(t)=f(\gamma(t))$$ and prove that g is differentiable for every t in R¹

Homework Equations

The Attempt at a Solution

I'm not having a strong understanding of differentiation of vector functions yet, so I'm not really sure if my proof is valid, check it please. Thanks!
f(x,y) is not differentiable at (0,0) so chain rule fails. If $$lim_{t->0} \frac{g(t)}{t} exists$$(since g(0)=0), it's done.
$$\frac{g(t)}{t}=\frac{f(\gamma(t))}{t}=\frac{\gamma^{3}_{1}(t)}{t(\gamma^{2}_{1}(t)+\gamma^{2}_{2}(t))}$$. when t approaches zero, the denominator is not zero since $$|\gamma'(0)|>0$$, the numerator can be applied the mean value theorem and becomes $$\gamma^{3}_{1}(t)-\gamma^{3}_{1}(0)=3\gamma^{2}_{1}(0)\gamma'_{1}(0)t=0\;since\;\gamma_{1}(0)=0$$. So g'(0) exists and equals to 0.

Thanks a lot!

actually...I'm a little more confident with it now

 

[boombaby]

okay now. Is this proof right?