$x^2y = 1 + cx$ → equation (1)
$x^2~dy + 2xy~dx = c~dx$
Divide by dx
$c = x^2~y' + 2xy$
Substitute c to equation (1)
$x^2y = 1 + (x^2~y' + 2xy)x$
$x^2y = 1 + x^3~y' + 2x^2y$
$1 + x^3~y' + x^2y = 0$
Multiply by dx
$dx + x^3~dy + x^2y~dx = 0$
$(1 + x^2y)~dx + x^3~dy = 0$ answer