I have been
"Use differentials to approximate the change in $f(x, y)$ as $(x, y)$ changes from $(−2, 3)$ to $(−2.02, 3.01)$."
$$f(x,y)=x^2-3x^3y^2+4x-2y^3+6$$
I have obtained:
$$\partial_xf(x,y)=2x-9x^2y^2+4,\qquad \partial_y f(x,y)=-6x^3-6y^2$$
Where do I go from here?
The total differential is the sum of the partials. So, $df = \partial_x f + \partial_y f$. Since $\partial_x f$ literally means the amount of changes in $f$ due to changes in $x$, then stick in the values for $x$, $y$, and $dx$ (i.e., the amount of change in $x$ from your starting point to your next point) into the formula for $\partial_x f$. Then, do the same thing for $\partial_y f$. Then add then together.