Answer
(a) Using the limit definition:
${f_x}\left( {0,0} \right) = {f_y}\left( {0,0} \right) = 0$
(b) Using the limit definition of directional derivative, we show that the directional derivative ${D_{\bf{u}}}f\left( {0,0} \right)$ does not exist for any unit vector ${\bf{u}}$ other than ${\bf{i}}$ and ${\bf{j}}$.
(c) By Theorem 2 of Section 15.4, $f\left( {x,y} \right)$ is not differentiable at $\left( {0,0} \right)$.
Work Step by Step
(a) We have $f\left( {x,y} \right) = {\left( {xy} \right)^{1/3}}$.
Evaluate ${f_x}\left( {0,0} \right)$ using the limit definition:
${f_x}\left( {0,0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {h,0} \right) - f\left( {0,0} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{0}{h} = 0$
Evaluate ${f_y}\left( {0,0} \right)$ using the limit definition:
${f_y}\left( {0,0} \right) = \mathop {\lim }\limits_{k \to 0} \frac{{f\left( {0,k} \right) - f\left( {0,0} \right)}}{k} = \mathop {\lim }\limits_{k \to 0} \frac{0}{k} = 0$
Hence, ${f_x}\left( {0,0} \right) = {f_y}\left( {0,0} \right) = 0$.
(b) Let ${\bf{u}} = \left( {h,k} \right)$.
Using the limit definition of directional derivative, we evaluate
${D_{\bf{u}}}f\left( {0,0} \right) = \mathop {\lim }\limits_{t \to 0} \frac{{f\left( {th,tk} \right) - f\left( {0,0} \right)}}{t}$
${D_{\bf{u}}}f\left( {0,0} \right) = \mathop {\lim }\limits_{t \to 0} \frac{{{{\left( {{t^2}hk} \right)}^{1/3}}}}{t} = \mathop {\lim }\limits_{t \to 0} \frac{{{{\left( {hk} \right)}^{1/3}}}}{{{t^{1/3}}}} = \infty $
Thus, ${D_{\bf{u}}}f\left( {0,0} \right)$ does not exist for any unit vector ${\bf{u}}$.
However, if we evaluate the directional derivative in the direction of ${\bf{i}}$ and ${\bf{j}}$:
${D_{\bf{i}}}f\left( {0,0} \right) = \mathop {\lim }\limits_{t \to 0} \frac{{f\left( {th,0} \right) - f\left( {0,0} \right)}}{t}$ ${\ \ }$ and ${\ \ }$ ${D_{\bf{j}}}f\left( {0,0} \right) = \mathop {\lim }\limits_{t \to 0} \frac{{f\left( {0,tk} \right) - f\left( {0,0} \right)}}{t}$
${D_{\bf{i}}}f\left( {0,0} \right) = \mathop {\lim }\limits_{t \to 0} \frac{0}{t} = 0$ ${\ \ }$ and ${\ \ }$ ${D_{\bf{j}}}f\left( {0,0} \right) = \mathop {\lim }\limits_{t \to 0} \frac{0}{t} = 0$,
we found that the limits exist.
Hence, the directional derivative ${D_{\bf{u}}}f\left( {0,0} \right)$ does not exist for any unit vector ${\bf{u}}$ other than ${\bf{i}}$ and ${\bf{j}}$.
(c) We evaluate the partial derivatives of $f\left( {x,y} \right) = {\left( {xy} \right)^{1/3}}$:
${f_x}\left( {x,y} \right) = \frac{{{y^{1/3}}}}{{3{x^{2/3}}}}$ ${\ \ }$ and ${\ \ }$ ${f_y}\left( {x,y} \right) = \frac{{{x^{1/3}}}}{{3{y^{2/3}}}}$
We notice that ${f_x}\left( {x,y} \right)$ and ${f_y}\left( {x,y} \right)$ do not exist at $\left( {0,0} \right)$. Hence, by Theorem 2 of Section 15.4, $f\left( {x,y} \right)$ is not differentiable at $\left( {0,0} \right)$.
