1. **State the problem:** Solve the equation $-8x^{-2} = 64x$ for $x$. 2. **Rewrite the equation:** Recall that $x^{-2} = \frac{1}{x^2}$. So the equation becomes: $$-8 \cdot \frac{1}{x^2} = 64x$$ 3. **Multiply both sides by $x^2$ to clear the denominator:** $$-8 \cancel{\cdot \frac{1}{x^2}} \cdot x^2 = 64x \cdot x^2$$ $$-8 = 64x^3$$ 4. **Isolate $x^3$:** $$x^3 = \frac{-8}{64}$$ 5. **Simplify the fraction:** $$x^3 = \frac{\cancel{-8}}{\cancel{64}} = -\frac{1}{8}$$ 6. **Solve for $x$ by taking the cube root:** $$x = \sqrt[3]{-\frac{1}{8}} = -\frac{1}{2}$$ **Final answer:** $$x = -\frac{1}{2}$$