<s>[INST] 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.
 [/INST] 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.
</s>