To find the equation of curve. We have given a differential equation we have to find the equation of curve. I tried it a lot but not able to proceed. Can someone give me some hints.
$(1-xy+x^2y^2)~dx=x^2~dy$
I tried to substitute $xy=t$ but got no result. Please help.

$(1-xy+x^2y^2)~dx=x^2~dy$
$\dfrac{dy}{dx}=\dfrac{1}{x^2}-\dfrac{y}{x}+y^2$
Let $y=-\dfrac{1}{u}$ ,
Then $\dfrac{dy}{dx}=\dfrac{1}{u^2}\dfrac{du}{dx}$
$\therefore\dfrac{1}{u^2}\dfrac{du}{dx}=\dfrac{1}{x^2}+\dfrac{1}{xu}+\dfrac{1}{u^2}$
$\dfrac{du}{dx}=\dfrac{u^2}{x^2}+\dfrac{u}{x}+1$
Let $v=\dfrac{u}{x}$ ,
Then $u=xv$
$\dfrac{du}{dx}=x\dfrac{dv}{dx}+v$
$\therefore x\dfrac{dv}{dx}+v=v^2+v+1$
$x\dfrac{dv}{dx}=v^2+1$
$\dfrac{dv}{v^2+1}=\dfrac{dx}{x}$
$\int\dfrac{dv}{v^2+1}=\int\dfrac{dx}{x}$
$\tan^{-1}v=\ln x+c$
$v=\tan(\ln x+c)$
$\dfrac{u}{x}=\tan(\ln x+c)$
$-\dfrac{1}{xy}=\tan(\ln x+c)$
$y=-\dfrac{\cot(\ln x+c)}{x}$
$y=\dfrac{\cot(-\ln x-c)}{x}$
$y=\dfrac{\tan(\ln x+C)}{x}$
