$\displaystyle 3x^2 \ln(1-xy)$
the partial derivatvie with respect to y is
$\displaystyle \frac{x}{1-xy}$
is that right?
Yes, it is treated like a constant.
What do you do to differentiate a constant times a function?
It's the constant times the derivative of that function.
i.e. $\displaystyle \frac{d}{dx}\left[c\cdot f(x)\right] = c\cdot f'(x)$.
So in your case, you have $\displaystyle 3x^2 \ln{(1 - xy)}$
The $\displaystyle 3x^2$ is your constant.
So the partial derivative will be
$\displaystyle 3x^2 \frac{\partial}{\partial y}\left[\ln{(1 - xy)}\right]$.