You can solve the problem by using the method of lagrange multiplier as you have already written in the comment.
$$\mathcal L=X^{1/3}\cdot Y^{1/2}+\lambda (I-P_x\cdot X-P_y\cdot Y)$$
Then you have to calculate the partial derivatives w.r.t $X,Y$ and $\lambda$.
$\frac{\partial \mathcal L}{\partial X}=\frac13\cdot X^{-2/3}\cdot Y^{1/2}-P_X\lambda=0 \Rightarrow\frac13\cdot X^{-2/3}\cdot Y^{1/2}=P_x\lambda \quad (1)$
$\frac{\partial \mathcal L}{\partial Y}=\frac12\cdot X^{1/3}\cdot Y^{-1/2}-P_y\lambda=0 \Rightarrow\frac12\cdot X^{1/3}\cdot Y^{-1/2}=P_y\lambda \quad (2)$
$\frac{\partial \mathcal L}{\partial \lambda}=I-P_x\cdot X-P_y\cdot Y=0 \quad (3)$
Dividing (1) by (2)
$\frac23\frac{Y}{X}=\frac{P_x}{P_y}$
Solving for $P_y\cdot Y$
$P_y\cdot Y=\frac32P_x\cdot X$. The expression can be insert in (3).
$I-P_x\cdot X-\frac32P_x\cdot X=0$
What is left is to solve the equation for $X$. Can you finish?