Given a function $f: \mathbb{R}^2 \to \mathbb{R}$ defined by $$f(x,y) = \begin{cases} \left( x^2 + y^2 \right) \cos \left(\frac{1}{x^2 + y^2} \right) & \text { if } (x,y) \neq (0,0)\\ 0 & \text{ if } (x,y) = (0,0) \end{cases}$$ compute the directional derivatives of $f$ at every point.
My solution:
$(x^2 + y^2)cos(\frac{1}{x^2 + y^2})$ at $(x,y) = (p,p)$ in the direction $\vec{v}=(\vec{v_1},\vec{v_2})$.
The partial derivatives are:
$\frac{(2p^2)cos(\frac{1}{2p^2})+sin(\frac{1}{2p^2})}{p}$ and $\frac{(2p^2)cos(\frac{1}{2p^2})+sin(\frac{1}{2p^2})}{p}$
The $\vec{v}$ becomes: $\frac{\vec{v_1}}{\sqrt{\vec{v_1^2} + \vec{v_2^2}}}$ and $\frac{\vec{v_2}}{\sqrt{\vec{v_1^2} + \vec{v_2^2}}}$
Then the product of the partial derivatives and the vectors is:
$D_\vec{v}f(p,p) = \frac{(v_1 + v_2)2p^2cos(\frac{1}{2p^2})+sin(\frac{1}{2p^2})}{p\sqrt{v_1^2 + v_2^2}}$
Am I correct?
You should not set $(x,y)=(p,p)$: for an arbitrary point on the plane $\mathbf{R}^2$, its $x$ and $y$ coordinates are not necessarily the same.
For $(x,y)\ne (0,0)$, you simply apply the formula $$ D_vf(x,y)=\nabla f(x,y)\cdot v=f_x(x,y)v_1+f_y(x,y)v_2. $$ where $v=(v_1,v_2)$ is a unit vector.
For $(x,y)= (0,0)$, work with the limit definition of partial derivatives.