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**ThePerfectHacker** Define the function $\displaystyle f:\mathbb{R}^2\mapsto \mathbb{R}$ as $\displaystyle f(x,y) = \left\{ \begin{array}{c} \tfrac{x^2y}{x^2+y^2} \mbox{ if }(x,y)\not = (0,0) \\ 0 \mbox{ if }(x,y)=0 \end{array} \right.$

The function is continous at $\displaystyle (0,0)$ as an added bonus. Because if $\displaystyle \epsilon > 0$ choose $\displaystyle \delta = \epsilon$ and so if $\displaystyle 0<\sqrt{x^2+y^2} < \delta$ it follows that $\displaystyle |f(x,y) | = \left| \tfrac{x^2y}{x^2+y^2} \right| \leq \left| \tfrac{x^2y}{2xy} \right| = \tfrac{1}{2}|x| \leq \tfrac{1}{2}\sqrt{x^2+y^2} < \epsilon$. (Here we used the inequality $\displaystyle x^2+y^2\geq 2xy$).

Next compute the directional derivatives $\displaystyle \partial_{\bold{u}}f(0,0)$ by definition. Now **if** the partial derivatives were to exist they would be related to the directional derivatives by $\displaystyle \partial_{\bold{u}}f(0,0) = \nabla f(0,0)\cdot \bold{u}$. But it does not. Therefore this function has directional derivatives but not partial.