## Calculus (3rd Edition)

$y'=\frac{y(y^{2}-x^{2})}{x(y^{2}-x^{2}-2xy^{2})}$
Differentiating $\frac{y}{x}+\frac{x}{y}=2y$ with respect to $x$, we have $\frac{y'\times x-y\times1}{x^{2}}+\frac{1\times y-x\times y'}{y^{2}}=2y'$ $\implies \frac{y^{2}(y'x-y)+(y-xy')x^{2}}{x^{2}y^{2}}=2y'$ $\implies y^{2}y'x-y^{3}+yx^{2}-x^{3}y'=2y'x^{2}y^{2}$ $\implies y^{2}y'x-x^{3}y'-2y'x^{2}y^{2}=y^{3}-yx^{2}$ $\implies y'(y^{2}x-x^{3}-2x^{2}y^{2})=y^{3}-yx^{2}$ $\implies y'= \frac{y^{3}-yx^{2}}{y^{2}x-x^{3}-2x^{2}y^{2}}=\frac{y(y^{2}-x^{2})}{x(y^{2}-x^{2}-2xy^{2})}$