Implicit Differentiation: $(x/y)+(y/x) =1$ Hi can anyone please tell me where I goes wrong with this question:
Find $ \frac{dy}{dx} $ for the curves defines by this equation:
\begin{align}
   \frac{x}{y} + \frac{y}{x} = 1 
 \end{align}
Here is what I did:
\begin{align}
    &\frac{y-xy'}{y^2} + \frac{y'x - y}{x^2} =0 \\
    &\frac{yx^2 - x^3 y'+y'y^2 x - y^3}{x^2 y^2} = 0 \\
    &yx^2 + (-x^3 +y^2 x)y' -y^3 =0 \\
    &\therefore y'= \frac{y^3 -yx^2}{-x^3 +y^2x}
\end{align}
The answer say it should be: $ y' = \frac{y}{x} $ but I had no clue how to proceed from there. 
Please help, Thanks.

we get by the quotient and the chaine rule
$$\frac{y-xy'}{y^2}+\frac{y'x-y}{x^2}=0$$ multiplying by $x^2y^2$ we obtain
$$x^2y-x^3y'+y'y^2x-y^3=0$$
solving for $y'$ we get
$$y'=\frac{x^2y-y^3}{x^3-xy^2}$$ if $x^3-y^2x\ne 0$
