Answer
$$y = 3{e^{2x - \tanh x}}$$
Work Step by Step
$$\eqalign{
& y'{\cosh ^2}x - y\cosh 2x = 0,\,\,\,\,\,\,\,y\left( 0 \right) = 3 \cr
& {\text{write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}}{\cosh ^2}x - y\cosh 2x = 0 \cr
& \cr
& {\text{separating the variables}} \cr
& \frac{{dy}}{{dx}}{\cosh ^2}x = y\cosh 2x \cr
& \frac{1}{y}dy = \frac{{\cosh 2x}}{{{{\cosh }^2}x}}dx \cr
& \cr
& {\text{Use the hyperbolic identity }}\cosh 2x = 2{\cosh ^2}x - 1 \cr
& \frac{1}{y}dy = \frac{{2{{\cosh }^2}x - 1}}{{{{\cosh }^2}x}}dx \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{1}{y}} dy = \int {\frac{{2{{\cosh }^2}x - 1}}{{{{\cosh }^2}x}}} dx \cr
& \int {\frac{1}{y}} dy = \int {\left( {\frac{{2{{\cosh }^2}x}}{{{{\cosh }^2}x}} - \frac{1}{{{{\cosh }^2}x}}} \right)} dx \cr
& \int {\frac{1}{y}} dy = \int {\left( {2 - {{\operatorname{sech} }^2}x} \right)} dx \cr
& \ln \left| y \right| = 2x - \tanh x + C \cr
& \cr
& {\text{use the initial condition }}y\left( 0 \right) = 3 \cr
& \ln \left| 3 \right| = 2\left( 0 \right) - \tanh \left( 0 \right) + C \cr
& \ln 3 = C \cr
& \cr
& Then,{\text{ the particular solution of the differential equation is}} \cr
& \ln \left| y \right| = 2x - \tanh x + \ln 3 \cr
& {\text{Solve the equation for }}y \cr
& {e^{\ln \left| y \right|}} = {e^{2x - \tanh x}}{e^{\ln 3}} \cr
& y = 3{e^{2x - \tanh x}} \cr} $$
