The equation of a curve passing through the origin and satisfying the differential equation $\frac{d y}{d x}=(x-y)^2$, is

$e^{2 x}(1-x+y)=1+x-y$

We have, $\frac{d y}{d x}=(x-y)^2$

Let $x-y=v$. Then,

$1-\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=1-\frac{d v}{d x}$

∴ $\frac{d y}{d x}=(x-y)^2$

$\Rightarrow 1-\frac{d v}{d x}=v^2$

$\Rightarrow 1-v^2 = \frac{d v}{d x}$

$\Rightarrow d x=\frac{1}{1-v^2} d v$

$\Rightarrow 2 \int d x=2 \int \frac{1}{1-v^2} d v$

$\Rightarrow 2 x=\log \left(\frac{1+v}{1-v}\right)+\log C$

$\Rightarrow C\left(\frac{1+v}{1-v}\right)=e^{2 x}$

$\Rightarrow C\left(\frac{x-y+1}{y-x+1}\right)=e^{2 x} \Rightarrow C(x-y+1)=e^{2 x}(y-x+1)$

Taking $C=1$, we find that option (a) is correct.