1. Given differential equation

   $\frac{dy}{dx}=(x-y)^2$ with initial condition $y(1)=1.$

2. Use a substitution

   Since the RHS contains $x-y$, let $u=x-y.$ Then $y=x-u.$ Differentiating w.r.t. $x$, $\frac{dy}{dx}=1-\frac{du}{dx}.$

   Substitute into the differential equation: $1-\frac{du}{dx}=u^2.$ So, $\frac{du}{dx}=1-u^2.$

3. Separate variables

   $\frac{du}{1-u^2}=dx.$

   Now, $\frac{1}{1-u^2}=\frac{1}{(1-u)(1+u)}.$ Hence, $\int \frac{du}{1-u^2}=\int dx.$

   Using the standard integral, $\int \frac{du}{1-u^2}=\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|.$

   Therefore, $\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|=x+C.$

   Multiply by 2: $\ln\left|\frac{1+u}{1-u}\right|=2x+C_1.$

4. Apply initial condition

   At $x=1$, $y=1$, so $u=x-y=1-1=0.$

   Substitute into the solution: $\ln\left|\frac{1+0}{1-0}\right|=2(1)+C_1$ $0=2+C_1$ $C_1=-2.$

   Thus, $\ln\left|\frac{1+u}{1-u}\right|=2x-2=2(x-1).$

5. Replace $u=x-y$

   $\ln\left|\frac{1+x-y}{1-x+y}\right|=2(x-1).$

   This is equivalent to $-\ln\left|\frac{1-x+y}{1+x-y}\right|=2(x-1).$

6. Compare with options

   Option D is $-\log_e\left|\frac{1-x+y}{1+x-y}\right|=2(x-1),$ which matches exactly.

Therefore, the correct answer is D.