I had a test last week and I failed it. Moreover, my teacher does not want to show the corrected test, and I don't know what I did wrong. So...
What are the partial derivatives of $f(x,y)=(xy)^{1/3}$ at $(0,0)$? Do partial derivatives exist at the points in the form $(a,0)$ and $(0,b)$, with $a,b$ nonzero? And $f$ is differentiable at $(0,0)$?
Thank you very much.
Hint: For $f_x(0,0)$, you need to compute
$$ \lim_{h\rightarrow 0}\frac{f(h,0)-f(0,0)}{h-0}=\lim_{h\rightarrow 0}\frac{(h\cdot 0)^{1/3}-(0\cdot 0)^{1/3}}{h}=\lim_{h\rightarrow 0}\frac{0}{h} $$ This looks like an indeterminate form and one could use l'Hopital's rule on it, but instead, observe that this is always zero (except when $h$ is zero), so this is the same as $\lim_{h\rightarrow 0}=0$.
For $f_y(a,0)$, you need to compute $$ \lim_{h\rightarrow 0}\frac{f(a,h)-f(a,0)}{h-a}=\lim_{h\rightarrow 0}\frac{(ah)^{1/3}-0}{h}=\lim_{h\rightarrow 0}\frac{a^{1/3}}{h^{2/3}}. $$ In this case, the denominator is going to zero while the numerator is constant. This limit diverges (to $\infty$).