I was sorely tempted to remark on your counting ability!

By the chain rule, $\displaystyle \frac{\partial w}{\partial x} = \frac{df}{du}\frac{\partial u}{\partial x}$ Since you are not given f you really just need to find the partial derivatives of u.

You want the "total differential": $\displaystyle Df= \frac{\partial f}{\partial y}dx+ \frac{\partial f}{\partial y}dy$= $\displaystyle 2xdx- 6ydy$

I recommend changing to polar coordinates. That way "closeness" to (0,0) is determined entirely by "r" and does not depend on "$\displaystyle \theta$".

$\displaystyle f(x,y)= \frac{(r cos(\theta))(r sin(\theta))}{r}= r cos(\theta)sin(\theta)$

If the limit of that, as r goes to 0, does not depend on $\displaystyle \theta$ then the limit of the function at (0,0) exists and is that value.