$ x\cos y + y \cos x = \pi$ , how do I find $y''(0)$ in easiest way possible? I considered brute force differentiating but that is very hard. I also tried expanding the cosines and '$y$' as a Taylor series but I  don't think that helps much either

$x\cos y + y\cos x = \pi$
What is $y(0)$?
$0\cos (y(0)) + y(0)\cos 0 = \pi\
y(0) = \pi\
What is $y'$?
$\cos y -y\sin x + (-x\sin y + \cos x)y' = 0\\
y'= \frac {y\sin x - \cos y}{\cos x - x\sin y} = \frac {u}{v}\\
$
And $y'(0)$?
$u(0) = \pi \sin 0 - \cos \pi = 1\\
v(0) = \cos 0 - 0\sin \pi = 1\\
y'(0) = 1$
$y'' = \frac {u'v - uv'}{v^2}\\
u' = y\cos x + (\sin x + \sin y) y'\\
u'(0) = \pi\\
v' = -sin x - sin y + (-x\cos y) y'\\
v'(0) = 0\\
y''(0) = \pi$
