I'm supposed to find the directional derivative (in all directions) at the origin of $\displaystyle\: f(x,y)= \frac{xy^6}{(x^4+y^8)}.$
I think the answer is supposed to be $0$, but when I take the partial derivatives I still have $0$ in the denominator so I'm not sure how to do it. Thanks.
First, let us compute gradient of function $\,f\left(x,y\right) = \dfrac{xy^6}{x^4 + y^8}:$
$$ \nabla f\left(x,y\right) = \begin{bmatrix}f_x\\f_y \end{bmatrix} % = \begin{bmatrix}f_x\\f_y \end{bmatrix} ,\qquad \left\lbrace\begin{aligned} f_x &= \frac{y^6\left(x^4+y^8\right) - 4x^4y^6}{\left(x^4+y^8\right)^2} = \frac{y^6\left(y^8-3x^4\right)}{\left(x^4+y^8\right)^2} \\ f_y &= \frac{6xy^5\left(x^4+y^8\right) - 8xy^{13}}{\left(x^4+y^8\right)^2} = \frac{2xy^5\left(3x^4 - y^8\right)}{\left(x^4+y^8\right)^2} \end{aligned}\right.\implies \\ \implies\nabla f\left(x,y\right) = \begin{bmatrix}f_x\\f_y \end{bmatrix} = \frac{\left.3x^4 - y^8\right.}{\left(x^4+y^8\right)^2} \begin{bmatrix}-y^6\\2xy^5 \end{bmatrix}$$
Thus, derivative of $\,f\left(x,y\right)\,$ along given vector $\,\vec{\sf v}=\begin{bmatrix}a\\b\end{bmatrix}$ can be computed as
$$ f_{\vec{\sf v}}\left(x,y\right) = \left\langle \nabla f, \vec{\sf v}\right\rangle = \frac{\big(2bx - a y\big)\left(3x^4 - y^8\right)y^5}{\left(x^4+y^8\right)^2} $$
Hope you can pick it from there. You may have to compute limit as $\,\left(x,y\right)\to\left(0,0\right).\,$