If $xy+\frac{x^2}{y}=x^3y+y$, then $\frac{dy}{dx}$ is equal to

$\frac{y(3x^2y^2-2x-y^2)}{xy^2-x^3y^2- x^2 - y^2}$

The correct answer is Option (2) → $\frac{y(3x^2y^2-2x-y^2)}{xy^2-x^3y^2- x^2 - y^2}$

Given equation: $xy + \frac{x^2}{y} = x^3y + y$

Differentiating both sides with respect to x using implicit differentiation:

Left side:

$\frac{d}{dx}\left(xy + \frac{x^2}{y}\right) = \frac{d}{dx}(xy) + \frac{d}{dx}\left(\frac{x^2}{y}\right)$

$\frac{d}{dx}(xy) = y + x \frac{dy}{dx}$

$\frac{d}{dx}\left(\frac{x^2}{y}\right) = \frac{2x \cdot y - x^2 \frac{dy}{dx}}{y^2} = \frac{2xy - x^2 \frac{dy}{dx}}{y^2}$

Left side derivative = $y + x \frac{dy}{dx} + \frac{2xy - x^2 \frac{dy}{dx}}{y^2}$

Right side:

$\frac{d}{dx}(x^3 y + y) = \frac{d}{dx}(x^3 y) + \frac{d}{dx}(y) = 3x^2 y + x^3 \frac{dy}{dx} + \frac{dy}{dx} = 3x^2 y + (x^3 + 1)\frac{dy}{dx}$

Equating derivatives:

$y + x \frac{dy}{dx} + \frac{2xy - x^2 \frac{dy}{dx}}{y^2} = 3x^2 y + (x^3 + 1)\frac{dy}{dx}$

Combine terms with $\frac{dy}{dx}$:

$x \frac{dy}{dx} - \frac{x^2}{y^2} \frac{dy}{dx} - (x^3 + 1) \frac{dy}{dx} = 3x^2 y - y - \frac{2xy}{y^2}$

$\frac{dy}{dx} \left(x - \frac{x^2}{y^2} - x^3 - 1 \right) = 3x^2 y - y - \frac{2x}{y}$

Therefore:

$\frac{dy}{dx} = \frac{3x^2 y - y - \frac{2x}{y}}{x - \frac{x^2}{y^2} - x^3 - 1}$

Answer: $\frac{dy}{dx} = \frac{y(3x^2y^2-2x-y^2)}{xy^2-x^3y^2- x^2 - y^2}$