Implicit Diff. check I have the expression ${\dfrac{x^2+y^2}{x+y}}=3$ and I wanna find $dy/dx$. Here's my approach:
$x^2+y^2=3x+3y$
$\implies (x^2-3x)+(y^2-3y)=0$
$\implies (2x-3)+\dfrac{dy}{dx}(2y-3)=0$
${\implies \dfrac{dy}{dx}=\dfrac{3-2x}{2y-3}}$
Wolfram Alpha gives me a very different answer. This seems to work (i.e, the tangent I found at $x=3$ from this looks right), but I'm not sure.

implicit differentiating of $\frac{x^2+y^2}{x+y}=3$ gives
$\frac{2x+2yy')(x+y)-(x^2+y^2)(1+y')}{(x^2+y^2)^2}=0$
plugging $x^2+y^2=3(x+y)$ in this term we get
$(2x+2yy')(x+y)-3(x+y)(1+y')=0$
with $x+y\ne 0$ we get $2x+2yy'-3-y'=0$ and this is your result
