Example D.0.1.
Define \(f \colon \R^2 \to \R\) by
\begin{equation*}
f(x,y)
=
\begin{cases}
\dfrac{x^3}{x^2+y^2}, & \langle x,y \rangle \ne \langle 0,0 \rangle, \\
0, & \langle x,y \rangle = \langle 0,0 \rangle.
\end{cases}
\end{equation*}
Show that every directional derivative of \(f\) at \(\langle 0,0 \rangle\) exists, but \(f\) is not differentiable there.
Solution.
Let \(\vu=\langle a,b \rangle\) be a unit vector. For \(t \ne 0\text{,}\)
\begin{equation*}
f(t\vu)
=
f(ta,tb)
=
\frac{t^3a^3}{t^2(a^2+b^2)}
=
t\,\frac{a^3}{a^2+b^2}.
\end{equation*}
Therefore
\begin{equation*}
D_{\vu}f(0,0)
=
\lim_{t \to 0}\frac{f(t\vu)-f(0,0)}{t}
=
\frac{a^3}{a^2+b^2}.
\end{equation*}
Since \(\vu\) is a unit vector, \(a^2+b^2=1\text{,}\) so
\begin{equation*}
D_{\vu}f(0,0)=a^3.
\end{equation*}
Thus every directional derivative exists at the origin. If \(f\) were differentiable at \(\langle 0,0 \rangle\text{,}\) then the map \(\vu \mapsto D_{\vu}f(0,0)\) would have to be linear in \(\vu\text{.}\) But
\begin{equation*}
D_{\langle 1,0 \rangle}f(0,0)=1,
\qquad
D_{\langle 0,1 \rangle}f(0,0)=0,
\end{equation*}
while
\begin{equation*}
D_{\frac{1}{\sqrt{2}}\langle 1,1 \rangle}f(0,0)
=
\left(\frac{1}{\sqrt{2}}\right)^3
=
\frac{1}{2\sqrt{2}}.
\end{equation*}
A linear map with the first two values above would send \(\frac{1}{\sqrt{2}}\langle 1,1 \rangle\) to \(\frac{1}{\sqrt{2}}\text{,}\) not to \(\frac{1}{2\sqrt{2}}\text{.}\) So \(\vu \mapsto D_{\vu}f(0,0)\) is not linear, and therefore \(f\) is not differentiable at the origin.
