You could find the derivatives of u and v separately, but that would require solving for u and v as functions of x and y. Can you do that?
I would use "implicit differentiation". For example, differentiating the first equation, $\displaystyle u^2v- u= x^3- 2y^3$, with respect to x, wq have $\displaystyle 2uv du/dx+ u^2 dv/dx- du/dx= 3x^2$. As "differentials in terms of the differentials of x and y" that would be $\displaystyle 2uv du+ u^2 dv= 3x^2 dx$. Differentiating with respect to y would give $\displaystyle 2uv du/dy- du/dy= 6y^2$ or, as "differentials", $\displaystyle 2uv dv- du= 6y^2 dy$.