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

The given equation is:

$xy + \dfrac{x^2}{y} = x^3y + y$

That $\dfrac{x^2}{y}$ will be messy to differentiate, so multiply both sides by $y$:

$xy^2 + x^2 = x^3y^2 + y^2 \quad \cdots (1)$

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Differentiating both sides of $(1)$ w.r.t. $x$:

Left side:

$\dfrac{d}{dx}(xy^2) + \dfrac{d}{dx}(x^2) = y^2 + 2xyy' + 2x$

Right side:

$\dfrac{d}{dx}(x^3y^2) + \dfrac{d}{dx}(y^2) = 3x^2y^2 + 2x^3yy' + 2yy'$

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Setting them equal:

$y^2 + 2xyy' + 2x = 3x^2y^2 + 2x^3yy' + 2yy'$

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Collecting all $y'$ terms on one side:

$2xyy' - 2x^3yy' - 2yy' = 3x^2y^2 - y^2 - 2x$

$y'(2xy - 2x^3y - 2y) = 3x^2y^2 - y^2 - 2x$

$y' = \dfrac{3x^2y^2 - y^2 - 2x}{2xy - 2x^3y - 2y}$

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$\boxed{\dfrac{dy}{dx} = \dfrac{y(3x^2y^2 - 2x - y^2)}{xy^2 - x^2y^3 - x^2 - y^2}}$