It suffices to prove that both sides are equal to $S(n)$, the number of pairs of positive integers $(x,y)$ such that $x^2y^3\le n$. For each positive integral value of $x$, if $x^2y^3\le n$, then $y\le\sqrt[3]{\tfrac n{x^2}}$. Thus, there are $\left\lfloor\sqrt[3]{\tfrac n{x^2}}\right\rfloor$ possible integer values of $y$. Doing a summation over $x$ gives us \[S(n)=\left\lfloor\sqrt[3]{\frac{n}{1^2}}\right\rfloor + \left\lfloor\sqrt[3]{\frac{n}{2^2}}\right\rfloor + \left\lfloor\sqrt[3]{\frac{n}{3^2}}\right\rfloor + \cdots.\] Similarly, for each positive integral value of $y$, we count the number of possibilities for positive integer $x$. Finally we get \[S(n)=\left\lfloor\sqrt[2]{\frac{n}{1^3}}\right\rfloor + \left\lfloor\sqrt[2]{\frac{n}{2^3}}\right\rfloor + \left\lfloor\sqrt[2]{\frac{n}{3^3}}\right\rfloor + \cdots.\]