The equation of a curve passing through the point $(-1,3)$, given that the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2 x}{y^2}$, is :

$y=\left(3 x^2+24\right)^{1 / 3}$

P(1,-3)

slope $\frac{d y}{d x}=\frac{2 x}{y^2}$      cross multiplying denominators

$\Rightarrow y^2 d y=2 x d x$

Integrating both sides

$\int y^2 d y=\int 2 x d x$

$\frac{y^3}{3}=\frac{2 x^2}{2}+C$

$\frac{y^3}{3}=x^2+C$

$y^3=3 x^2+3 C$

$y^3=3 x^2+C'$        ....(1)

(Let 3C = C'  another constant)

curve passes through (1, -3)

putting value in equation of curve eq (1)

$(-3)^3=3(1)^2+e^1$

$\Rightarrow 27=3+C'$

$C'=24$

from (1)

$y^3=3 x^2+24$

$y=\left(3 x^2+24\right)^{1 / 3}$