Derivative of $\frac { y }{ x } +\frac { x }{ y } =2y$ with respect to $x$ $$\frac { y }{ x } +\frac { x }{ y } =2y$$
Steps I took:
$$\frac { d }{ dx } \left[yx^{ -1 }1+xy^{ -1 }\right]=\frac { d }{ dx } [2y]$$
$$\frac { dy }{ dx } \left(\frac { 1 }{ x } \right)+(y)\left(-\frac { 1 }{ x^{ 2 } } \right)+(1)\left(\frac { 1 }{ y } \right)+(x)\left(-\frac { 1 }{ y^{ 2 } } \right)\frac { dy }{ dx } =(2)\frac { dy }{ dx } $$
$$-\frac { y }{ x^{ 2 } } +\frac { 1 }{ y } =(2)\frac { dy }{ dx } -\left(\frac { 1 }{ x } \right)\frac { dy }{ dx } +\left(\frac { x }{ y^{ 2 } } \right)\frac { dy }{ dx } $$
$$-\frac { y }{ x^{ 2 } } +\frac { 1 }{ y } =\left(2-\frac { 1 }{ x } +\frac { x }{ y^{ 2 } } \right)\frac { dy }{ dx } $$
$$\frac { -\frac { y }{ x^{ 2 } } +\frac { 1 }{ y }  }{ \left(2-\frac { 1 }{ x } +\frac { x }{ y^{ 2 } } \right) } =\frac { dy }{ dx } $$
At this point I get stuck because once I simplify the result of the last step I took, the answer is not what it should be. I think that I am making a careless mistake somewhere but I cannot seem to find it. Hints only, please. The direct answer does nothing for me. 
Actual answer:
$$\frac { d y}{ dx } =\frac { y(y^{ 2 }-x^{ 2 }) }{ x(y^{ 2 }-x^{ 2 }-2xy^{ 2 }) } $$

it will be a whole lot easier to turn $$  \frac { y }{ x } +\frac { x }{ y } =2y$$ into $$ x^2 + y^2 = 2xy^2$$ and difference as $$2x\,dx + 2y\, dy = 4xy \, dy +2y^2 \, dx $$ which can then be simplifies as $$\frac{dy}{dx} = \frac{y^2 - x}{y(1-2x)}\, \text{ and subject to the onstraint  }  \frac { y }{ x } +\frac { x }{ y } =2y.$$
