<s>[INST] We have $(x^2+y^2)^2-3(x^2+y^2)+1=0$. What is the value of $\frac{d^2y}{dx^2}$? 
We have $(x^2+y^2)^2-3(x^2+y^2)+1=0$. What is the value of
$\frac{d^2y}{dx^2}$?
$1)-\frac{x^2+y^2}{y^2}\qquad\qquad2)-\frac{x^2+y^2}{y^3}\qquad\qquad3)\frac{x+y}{x^2+y^2}\qquad\qquad4)\frac{xy}{x^2+y^2}$

Here is my approach:
We have a quadratic equation in $x^2+y^2$. So $x^2+y^2=\frac{3\pm\sqrt5}{2}$ (RHS is a constant). Taking differentiate with respect to $x$,
$$2x+2yy'=0\Rightarrow\quad y'=\frac{-x}{y}$$
$$2+2y'^2+2yy''=0\Rightarrow \quad y''=\frac{-y'^2}{y}\Rightarrow y''=\frac{-x^2}{y^3}$$
Edit:
The second option can be written as $-(\frac{x^2}{y^3}+\frac{1}y)$. It seems it is the correct answer. But why I missed $-\frac1y$ in this approach?
 [/INST] Your approach is correct, you just made a small mistake at the last step: from $2+2y'^2+2yy''=0$ we find that
$$y''=-\frac{1+y'^2}{y}=-\frac{1+(-x/y)^2}{y}=-\frac{x^2+y^2}{y^3}.$$
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