For the equation given, evaluate y′ at the point (−2,−1) I must find the derivative for this equation
$$
   (3x−y)^4+4y^3=621
$$
 at the point $(-2,-1)$.
With some differential calculus: differentiating both sides you obtain:
$$4(3x-y)^3(3\,\mathrm dx -\mathrm dy)+12 y^2\,\mathrm dy=0$$
which becomes for $x=-2,\;y=-1$:
$$-500(3\,\mathrm dx -\mathrm dy)+12 \,\mathrm dy=0\iff 512\,\mathrm dy=1500\,\mathrm dx\iff\frac{\mathrm dy}{\mathrm dx}=\frac{1500}{512}=\frac{375}{128}.$$
Use $$\left((3x-y)^4+4y^3\right)'=0.$$

We obtain $$4(3x-y)^3(3-y')+12y^2y'=0.$$
Can you end it now?