The ''derivative'' of a vector function of a vector is not a single number, but a matrix that contains all the partial derivative of the vector function with respect to the components of the independent vector, called the Jacobian matrix.
In your case, supposing that, $\vec W$ and $\vec v$ are vectors of a $n-$ dimensional real vector space, we have:
$$
\frac{\partial \vec w}{\partial \vec v}=
\begin{bmatrix}
\frac{\partial w_1}{\partial v_1}&\frac{\partial w_1}{\partial v_2} &\cdots&\frac{\partial w_1}{\partial v_n}\\
\frac{\partial w_2}{\partial v_1}&\frac{\partial w_2}{\partial v_2} &\cdots&\frac{\partial w_2}{\partial v_n}\\
\cdots\\
\frac{\partial w_n}{\partial v_1}&\frac{\partial w_n}{\partial v_2} &\cdots&\frac{\partial w_n}{\partial v_n}\\
\end{bmatrix}
$$
Using this definition you can evaluate the elements of the matrix for your function. If $n=2$, we have:
$$
\begin{bmatrix}
w_1\\
w_2
\end{bmatrix}
=
\begin{bmatrix}
\frac{gv_1}{\sqrt{v_1^2+v_2^2}}\\
\frac{gv_2}{\sqrt{v_1^2+v_2^2}}
\end{bmatrix}
$$
so we can find the partial derivatives as:
$$
\frac{\partial w_1}{\partial v_1}=\frac{gv_2^2}{\sqrt{\left(v_1^2+v_2^2\right)^3}}
\qquad
\frac{\partial w_1}{\partial v_2}=\frac{-gv_1v_2}{\sqrt{\left(v_1^2+v_2^2\right)^3}}
$$
and so one... and you can do the same if $n>2$.
