The solution of the differential equation,  $\frac{dy}{dx}=(x-y{)}^2,$ when $y\left(1\right)=1$ is 

Given, $\frac{dy}{dx}=(x-y{)}^2$ 

Let, $x-y=t\Rightarrow \frac{dy}{dx}=1-\frac{dt}{dx}\therefore 1-\frac{dt}{dx}=t^2$

$\begin{array}{l}\int \frac{dt}{1-t^2}=\int dx\Rightarrow \frac{1}{2}{\log }_e\left|\frac{1+t}{1-t}\right|=x+c\\ \frac{1}{2}{\log }_e\left|\frac{1+x-y}{1-x+y}\right|=x+C\end{array}$

Given, $y\left(1\right)=1$

$\begin{array}{l}\therefore \frac{1}{2}{\log }_e\left(1\right)=C+1\Rightarrow C=-1\\ \therefore {\log }_e\left|\frac{1+x-y}{1-x+y}\right|=2(x-1)\Rightarrow -{\log }_e\left|\frac{1-x+y}{1+x-y}\right|=2(x-1)\end{array}$