This is a first-order linear differential equation in $x$ of the form $\frac{dx}{dy} + P(y)x = Q(y)$:
Here $P(y) = \frac{1}{y}$ and $Q(y) = y^2$.
Integrating Factor (I.F.):

$$\text{I.F.} = e^{\int P(y)\, dy} = e^{\int \frac{1}{y}\, dy} = e^{\ln y} = y$$

The general solution is given by:

$$x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.})\, dy + c$$ $$x \cdot y = \int y^2 \cdot y\, dy + c = \int y^3\, dy + c = \frac{y^4}{4} + c$$

Dividing throughout by $y$:

$$x = \frac{1}{4}y^3 + \frac{c}{y}$$