y = √x is not a solution because y = √x has derivative also involving √x, and rational powers won't get resolved in the ODE. That is,
[tex]y = \sqrt x \implies y' = \dfrac1{2\sqrt x} \implies xy'+y = \dfrac x{2\sqrt x}+\sqrt x = \dfrac32\sqrt x \neq x^2[/tex]
Similarly, y = 5/x is not a solution because its derivative is a rational function that also doesn't get resolved.
[tex]y=\dfrac5x \implies y=-\dfrac5{x^2} \implies xy'+y = -\dfrac5x+\dfrac5x=0\neq x^2[/tex]
In the same vein, y = sin(x) has derivative y' = cos(x), and these trigonometric expression won't get resolved in this case.
[tex]y=\sin(x)\implies y'=\cos(x) \implies xy'+y=x\cos(x)+\sin(x) \neq x^2[/tex]
So we focus on the remaining candidate:
[tex]y = \dfrac{x^2}3 \implies y' = \dfrac{2x}3 \implies xy'+y = \dfrac{2x^2}3+\dfrac{x^2}3 = x^2[/tex]
and y = x²/3 is the correct choice.
