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