Answer
\[y(x)=\frac{1}{3}\left[3x-\tan^{-1}(3x+C)+1\right]\]
Work Step by Step
\[y'=\sin^2(3x-3y+1)\;\;\;\ldots (1)\]
Substitute $\: V=3x-3y+1 \;\;\;\ldots (2) $
\[\frac{dV}{dx}=3-3\frac{dy}{dx}\]
From (1) and (2)
\[\frac{dV}{dx}=3-3\left(\sin^2 V\right)=3\left(1-\sin^2 V\right)\]
\[\frac{dV}{dx}=3\cos^2 V\]
Separating variables
\[\sec^2 VdV=3dx\]
Integrating , \[\int\sec^2 VdV=3\int dx+C\]
$C$ is constant of integration
\[\tan V=3x+C\]
\[V=\tan^{-1}(3x+C)\]
From (2)
\[3x-3y+1=\tan^{-1}(3x+C)\]
\[y=\frac{1}{3}\left[3x-\tan^{-1}(3x+C)+1\right]\]
Hence \[y(x)=\frac{1}{3}\left[3x-\tan^{-1}(3x+C)+1\right].\]
