From question, the differential equation given is:

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

The general form of the given equation is:

\(\frac{{dy}}{{dx}} = f\left( {ax + by + c} \right)\)

On putting, x - y = t

\(1 - \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}} \Rightarrow \frac{{dy}}{{dx}} = 1 - \frac{{dt}}{{dx}}\)

Now, the equation becomes,

\(\Rightarrow 1 - \frac{{dt}}{{dx}} = {t^2}\)

\(\Rightarrow \frac{{dt}}{{dx}} = 1 - {t^2}\)

On separating the variables,

\(\Rightarrow \smallint \frac{{dt}}{{1 - {t^2}}} = \smallint dx\)

∵ \(\left[ {\smallint \frac{{dx}}{{{a^2} - {x^2}}} = \frac{1}{{2a}}{\rm{lo}}{{\rm{g}}_e}\left| {\frac{{a + x}}{{a - x}}} \right| + C} \right]\)

\(\Rightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_e}\left( {\frac{{1 + t}}{{1 - t}}} \right) = x + C\)

\(\Rightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_e}\left( {\frac{{1 + x - y}}{{1 - x + y}}} \right) = x + C\)

From question, when x = 1 then y = 1,

\(\Rightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_e}\left( {\frac{{1 + 0}}{{1 + 0}}} \right) = 1 + C\)

\(\Rightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_e}\left( {\frac{{1 + x - y}}{{1 - x + y}}} \right) = x - 1\)

∵ \(\left[ {{\rm{log}}\frac{1}{x} = {\rm{log}}{x^{ - 1}} = - {\rm{log}}x} \right]\)