Problem: Solve for $x$ given $x^{-1/3} = y$. 1. Observe the expression and rewrite the power in root form to understand it better. $x^{-1/3} = 1/\sqrt[3]{x}$. 2. To isolate $x$, raise both sides of the equation $x^{-1/3} = y$ to the power $-3$. 3. Apply the exponent rule $ (a^{m})^{n} = a^{mn} $ so that $(x^{-1/3})^{-3} = x^{(-1/3)(-3)} = x^{1} = x$ and the right-hand side becomes $y^{-3}$. 4. Combine the results to get the solution. $x = y^{-3}$. 5. Check domain and special cases in plain language: $x^{-1/3}=1/\sqrt[3]{x}$ is undefined at $x=0$, so the original equation can never equal $0$, hence $y\neq 0$. Therefore for any nonzero real $y$ the solution is $x = y^{-3}$ and $x\neq 0$. Final answer: $x = y^{-3}$ for $y \neq 0$.