If x*cos(y)+y*cos(x) = pi, what is y''(0)?

Level: Advanced

Applying implicit differentiation to the given equation,
(1) cos(y)-x*sin(y)*y'+y'*cos(x)-y*sin(x) = 0.
Apply implicit differentiation to (1),
(2) -sin(y)*y'-x[cos(y)*y'+sin(y)*y'']+y''*cos(x)-y'*sin(x)-y'*sin(x)-y*cos(x) = 0.
Setting x = 0 in the given equation,
0+y(0)*1 = pi,
y(0) = pi.
Setting x = 0 and y(0) = pi in (1),
-1-0+y'(0)*1-0 = 0,
y'(0) = 1.
Setting x = 0, y(0) = pi, and y'(0) = 1 in (2),
-0-0+y''(0)*1-0-0-pi*1 = 0,
y''(0) = pi.