<s>[INST] Solve $y' + y^2 = \frac{1}{x^2}$ by introducing $z = xy$ as a new function. Question:
Solve the equation:
$$y' + y^2 = \frac{1}{x^2},~~~~~~~~~x > 0$$
by introducing $z = xy$
Attempted answer:
$z = xy \Rightarrow y = \frac{z}{x}$
Taking the derivative of $y$ with respect to $x$ using he product rule:
$$y' = \frac{z'}{x} - \frac{z}{x}$$
Adding this into the original equation:
$$\frac{z'}{x} - \frac{z}{x^2} + \frac{z^2}{x^2} = \frac{1}{x^2}$$
Putting $z$ and $x$ on each side:
$$\frac{z'}{1-z^2 + z} = \frac{1}{x}$$
Integrating on each side:
$$\int \frac{1}{1-z^2 + z}dz = \int \frac{1}{x} dx$$
This produces
$$z(x) = \frac{(1+\sqrt{5})c_1x^{\sqrt{5}}+1-\sqrt{5}}{(2c_1 x^{\sqrt{5}}+2)}$$
Substituting back to $y$:
$$y(x) = \frac{(1+\sqrt{5})c_1x^{\sqrt{5}}+1-\sqrt{5}}{x(2c_1 x^{\sqrt{5}}+2)}$$
This seems all and well, but the expected answer contains yet another solution:
$$y = \frac{1\pm\sqrt{5}}{2x}$$
How does this second solution arise? Since there is an $x^2$ in the question, I think that a solution might have been dropped at some point.
 [/INST] Because of the division that must be different from zero as GSofer pointed in his good answer but you can also deduce that solution by other methods:
$$y' + y^2 = \frac{1}{x^2}$$
By inspection $y_p=\frac  A x$
$$-\frac A {x^2}+\frac {A^2}{x^2}=\frac 1 {x^2}$$
$$ \implies A^2-A-1=0$$
$$\implies A=\frac 12 \pm\frac {\sqrt 5}{2} \implies y =\frac {1\pm \sqrt 5}{2x}$$
</s>