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

$\text{ Assuming your y' is correct... } \\ \text{  then we should get rid of the compound fractions.. } \\ y'=\frac{\frac{-y}{x^2}+\frac{1}{y}}{2-\frac{1}{x}+\frac{x}{y^2}} \\  \text{ now we need to multiply top and bottom by } \\ x^2y^2 \text{ this is the lcm of the bottoms of the mini-fractions } \\ y'=\frac{-y(y^2)+1(x^2)(y)}{2x^2y^2-xy^2+x(x^2)} \\ y'=\frac{-y^3+x^2y}{2x^2y^2-xy^2+x^3}$
