Answer
(a) Using the limit definition of directional derivative we show that:
${D_{\bf{v}}}f\left( {0,0} \right)$ exists for all vectors ${\bf{v}}$.
Using the limit definition we show that:
${f_x}\left( {0,0} \right) = {f_y}\left( {0,0} \right) = 0$
(b) We show that $f$ does not satisfy Eq. (8). Hence, $f$ is not differentiable at $\left( {0,0} \right)$.
Work Step by Step
(a) We have the definition:
$f\left( {x,y} \right) = \frac{{{x^2}y}}{{{x^2} + {y^2}}}$ ${\ \ }$ for $\left( {x,y} \right) \ne \left( {0,0} \right)$ and $f\left( {0,0} \right) = 0$.
Let ${\bf{v}} = \left( {h,k} \right)$.
1. Using the limit definition of directional derivative, we evaluate
${D_{\bf{v}}}f\left( {0,0} \right) = \mathop {\lim }\limits_{t \to 0} \frac{{f\left( {th,tk} \right) - f\left( {0,0} \right)}}{t}$
${D_{\bf{v}}}f\left( {0,0} \right) = \mathop {\lim }\limits_{t \to 0} \frac{{{t^3}{h^2}k/\left( {{t^2}{h^2} + {t^2}{k^2}} \right)}}{t}$
$ = \mathop {\lim }\limits_{t \to 0} \frac{{{h^2}k}}{{{h^2} + {k^2}}} = \frac{{{h^2}k}}{{{h^2} + {k^2}}}$
Thus, ${D_{\bf{v}}}f\left( {0,0} \right)$ exists for all vectors ${\bf{v}}$.
2. 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) Recall from Exercise 69: if $f\left( {x,y} \right)$ is differentiable at $\left( {0,0} \right)$ and
$f\left( {0,0} \right) = {f_x}\left( {0,0} \right) = {f_y}\left( {0,0} \right) = 0$,
then Eq. (8) holds:
(8) ${\ \ \ }$ $\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{f\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }} = 0$
Using this result, we check the differentiability of $f$ at $\left( {0,0} \right)$ by evaluating the limit:
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{f\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^2}y}}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}}.$
Let us choose the case when we approach $\left( {0,0} \right)$ along the line $y=x$. So, substituting it in the limit gives
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{f\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^2}y}}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}}$
$ = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^3}}}{{{{\left( {2{x^2}} \right)}^{3/2}}}}$
$ = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{1}{{{{\left( 2 \right)}^{3/2}}}} = \frac{1}{{2\sqrt 2 }}$
But the limit is not zero. Therefore, $f$ does not satisfy Eq. (8). Hence, $f$ is not differentiable at $\left( {0,0} \right)$.
