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

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

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

CASE I: When $\frac{d y}{d x}=(x-y)$

In this case, we have

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

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

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

$\Rightarrow -\log (1-v)=x+\log C$

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

It passes through the origin.

∴  $C=1$

Hence, $\frac{1}{1-x+y}=e^x$           ......(i)

CASE II: When $\frac{d y}{d x}=-(x-y)=y-x$

In this case, we have

$\frac{d u}{d x}+1=u$, where $u=y-x$

$\Rightarrow \frac{d u}{u-1}=d x$

$\Rightarrow \log (u-1)=x+\log C$

$\Rightarrow u-1=C e^x \Rightarrow y-x-1=C e^x$

It passes through the origin.

∴  $C=-1$

Hence, $y-x-1=-e^x$ or, $x-y+1=e^x$              ......(i)

From (i) and (ii), we obtain

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