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.
 A: if you going to do implicit differentiation you might as well multiply by $xy$ to get $x^2 + y^2 = xy$ before differencing. now differencing gives you $2xdx + 2ydy = xdy + ydx.$ this can also be written as 
$$\dfrac{dy}{dx} = \dfrac{y - 2x}{2y-x} = \dfrac{y(y-2x)}{y(2y-x)}=
\dfrac{y(y-2x)}{2y^2 - xy} = \dfrac{y(y-2x)}{2xy - 2x^2 - xy} = \dfrac{y(y-2x)}{x(y-2x)} = \dfrac{y}{x} \tag 1$$ 
in implicit differentiation you don't have a unique answer. you always have to carry the constraint $\dfrac{x}{y} + \dfrac{y}{x} = 1$ along or an equivalent one like $x^2 + y^2 = xy$ along with the solution. in the equality (1) any one of them could be an answer.
A: Note that $y=y(x)$
So we have that 
$$\frac{x}{y(x)} + \frac{y(x)}{x} = 1$$
So,
$$\begin{align} 0 &= \frac{d}{dx}\left( \frac{x}{y(x)} + \frac{y(x)}{x}\right) \\
&=\frac{d}{dx}(x) \frac{1}{y(x)}+x \frac{d}{dx}\left(\frac{1}{y(x)}\right) \\
&= \frac{1}{y(x)}+ x \ \frac{-1}{y^2(x)} \frac{dy}{dx} \\
&= \frac{1}{y(x)}\left(1-\frac{x}{y(x)} \right)\frac{dy}{dx}  
\end{align}$$
Assuming $\frac{1}{y(x)} \neq 0$ then
$$0= \left(1-\frac{x}{y(x)} \right)\frac{dy}{dx}   $$
$$-1= -\frac{x}{y(x)} \frac{dy}{dx} $$
$$\frac{dy}{dx}=\frac{y(x)}{x}=\frac{y}{x} $$
The only thing I used is the product rule http://en.wikipedia.org/wiki/Product_rule
A: 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 $x^3-y^2x\ne 0$
