Answer
True
Work Step by Step
If
\[
\cos (y+x)+\ln (3 x-3 y)=w
\]
and then
\[
\frac{1}{3 x-3 y} \cdot 3-\sin (x+y)=\frac{3}{3 x-3 y}-\sin (y+x)=w_{x}
\]
,
\[
\begin{aligned}
\frac{-3}{(3 x-3 y)^{2}} & \cdot 3-\cos (x+y)=\frac{-9}{9(x-y)^{2}}-\cos (x+y)=w_{x x} \\
&=\frac{-1}{(x-y)^{2}}-\cos (x+y)
\end{aligned}
\]
whereas
\[
w_{y}=-\sin (x+y)+\frac{1}{3 x-3 y} \cdot(-3)=\frac{-3}{3 x-3 y}-\sin (x+y)
\]
and
\[
\begin{array}{c}
\frac{-(-3)}{(3 x-3 y)^{2}} \cdot(-3)-\cos (x+y)=\frac{-9}{9(x-y)^{2}}-\cos (x+y)=w_{y y} \\
=-\cos (y+x)+\frac{-1}{(x-y)^{2}}
\end{array}
\]
