Answer
The tangent line at $x=\pi$ is $y=-x.$
Work Step by Step
The equation for the tangent line at $x_0$ is:
$y-f(x_0)=f'(x_0)(x-x_0)$
Here $x_0=\pi$ and $f(x)=y=x\cos3x$.
$$f'(x)=(x\cos3x)'=x'\cos3x+x(\cos 3x)'=1\cdot\cos3x+x(-\sin 3x)(3x)'=\cos3x-3x\sin 3x$$
Now for $x_0=\pi$ we have:
$$f(\pi)=\pi\cos3\pi=-\pi, f'(\pi)=\cos3\pi-3\pi\sin3\pi=-1-3\pi\cdot0=-1$$
The equation for the tangent line at $x=\pi$ is:
$$y-(-\pi)=-1\cdot(x-\pi)\Rightarrow y=-x$$
