If the slope of the tangent to the curve $y = y(x)$ at any point $(x, y)$ is $\frac{2x}{y^2}$ and the curve passes through the point $(\frac{1}{\sqrt{3}},1)$, then equation of curve is

$y^3 = 3x^2$

The correct answer is Option (2) → $y^3 = 3x^2$

Given: Slope of tangent $\frac{dy}{dx} = \frac{2x}{y^2}$, passes through $\left(\frac{1}{\sqrt{3}}, 1\right)$

Separate variables:

$y^2 dy = 2x dx$

Integrate both sides:

$\int y^2 dy = \int 2x dx$

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

Use point $(x, y) = (\frac{1}{\sqrt{3}}, 1)$ to find C:

$\frac{1^3}{3} = (\frac{1}{\sqrt{3}})^2 + C \Rightarrow \frac{1}{3} = \frac{1}{3} + C \Rightarrow C = 0$

Equation of curve:

$y^3 = 3x^2$