Step 1:

Given :The square of the difference of the abscissa and the ordinate is equal to the slope of the tangent of the curve.

Slope of the tangent to the curve is $\large\frac{dy}{dx}$

The square of the difference of the abscissa and the ordinate is $(x-y)^2$

Hence $\large\frac{dy}{dx}$$=(x-y)^2$

Put $x-y=z$ on differentiating both sides w.r.t x we get,

$1-\large\frac{dy}{dx}=\large\frac{dz}{dx}=>\frac{dy}{dx}$$=1-\large\frac{dz}{dx}$

Now substituting this we get,

$1-\large\frac{dz}{dx}$$=z^2$

rearranging this we get,

$\large\frac{dz}{dx}$$=1-z^2$

Now seperating the variables we get,

$\large\frac{dz}{1-z^2}$$=dx$

Step 2:

Integrating on both sides,

$\large \int \frac{dz}{1-z^2}$$=\int dx$

$\large \int \frac{dz}{1-z^2}$ is of the form $\large \int \frac{dx}{a^2-x^2}=\frac{1}{2a} $$\log \large\frac{|a+x|}{|a-x|}$$+c$

Here $a=1$ and $x=z$

$\int \large\frac{dz}{1-z^2}=\frac{1}{2} $$\log \bigg|\large\frac{1+z}{1-z}\bigg|$

Therefore $\large \frac{1}{2} $$\log \bigg|\large\frac{1+z}{1-z}\bigg|$$=x+c$

$\log_e \bigg|\large\frac{1+z}{1-z}\bigg|$$=2x+c$

Substituting for $z=x-y$

$\log_e \bigg|\large\frac{1+(x-y)}{1-x+y}\bigg|$$=2x+c$

$ce^{2x}=\large\frac{1+(x-y)}{(1-x+y)}$

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

It is given the curve passes through the orgin. Hence substitute $x=0$ and $y=0$ to evaluate the value of c

$ce^0(1-0+0)=1+(0-0)$

But $e^0=1$

$c=1$

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

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