1. **State the problem:** We need to find $\frac{dy}{dx}$ at the point $(2,0)$ for the implicit equation $$\tan^{-1}(x^2 y) = 4x + xy - 8.$$\n\n2. **Recall the formula and rules:** For implicit differentiation, differentiate both sides with respect to $x$, remembering to use the chain rule and product rule where necessary. Also, $\frac{d}{dx}[\tan^{-1}(u)] = \frac{1}{1+u^2} \cdot \frac{du}{dx}$.\n\n3. **Differentiate the left side:** Let $u = x^2 y$, then\n$$\frac{d}{dx}[\tan^{-1}(x^2 y)] = \frac{1}{1+(x^2 y)^2} \cdot \frac{d}{dx}[x^2 y].$$\nUsing product rule on $x^2 y$:\n$$\frac{d}{dx}[x^2 y] = 2x y + x^2 \frac{dy}{dx}.$$\nSo left side derivative is\n$$\frac{1}{1+x^4 y^2} (2x y + x^2 \frac{dy}{dx}).$$\n\n4. **Differentiate the right side:**\n$$\frac{d}{dx}[4x + xy - 8] = 4 + y + x \frac{dy}{dx} - 0 = 4 + y + x \frac{dy}{dx}.$$\n\n5. **Set derivatives equal:**\n$$\frac{1}{1+x^4 y^2} (2x y + x^2 \frac{dy}{dx}) = 4 + y + x \frac{dy}{dx}.$$\n\n6. **Plug in the point $(2,0)$:** Note $y=0$, so $x^4 y^2 = 2^4 \cdot 0 = 0$.\nLeft side becomes\n$$\frac{1}{1+0} (2 \cdot 2 \cdot 0 + 2^2 \frac{dy}{dx}) = 4 \frac{dy}{dx}.$$\nRight side becomes\n$$4 + 0 + 2 \frac{dy}{dx} = 4 + 2 \frac{dy}{dx}.$$\n\n7. **Equation at the point:**\n$$4 \frac{dy}{dx} = 4 + 2 \frac{dy}{dx}.$$\n\n8. **Solve for $\frac{dy}{dx}$:**\n$$4 \frac{dy}{dx} - 2 \frac{dy}{dx} = 4$$\n$$2 \frac{dy}{dx} = 4$$\n$$\frac{dy}{dx} = 2.$$\n\n**Final answer:** $\boxed{2}$