What is dy/dt when x=π/6, y=π/6, dx/dt=1

4xcos(y) - πtan(x) = 0

dy/dt = ?

4x cosy − π tanx = 0

First we differentiate implicitly to find dy/dx

4 cos y − 4x siny dy/dt − π sec²x = 0
−4x siny dy/dt = −(4 cos y − π sec²x)
dy/dt = (4 cos y − π sec²x) / (4x siny)

When x = π/6, y = π/6
dy/dx = (4 cos(π/6) - π sec^2(π/6)) / (4 (π/6) sin(π/6)) = 6√3/π − 4

By chain rule:
dy/dt = dy/dx * dx/dt
dy/dt = (6√3/π − 4) * 1

dy/dt = 6√3/π − 4

4x∙cos(y) - π∙tan(x) = 0

Differentiate Implicitly:
4(dx/dt)∙cos(y) - 4x∙sin(y)∙(dy/dt) - π∙sec²(x)∙(dx/dt) = 0

Substitute:
4(1)∙cos(π/6) - 4(π/6)∙sin(π/6)∙(dy/dt) - π∙sec²(π/6)∙(1) = 0
(2π/3)∙(√(3)/2)∙(dy/dt) = 2 - 4π
dy/dt = (2 - 4π)√(3)/π

d/dt(4xcosy - pitanx) = d/dt(0)
4cosy(dx/dt) + 4x(-siny)(dy/dt) - pisec^2x(dx/dt) = 0

4cos(pi/6)(1) + 4(pi/6)(-sin(pi/6))dy/dt - pisec^2(pi/6)(1) = 0
4(√3/2) + 2pi/3(-1/2)dy/dt - pi(1/(√3/2))^2 = 0
2√3 - pi/3(dy/dt) - 4pi/3 = 0
2√3 - 4pi/3 = pi/3(dy/dt)
6√3/pi - 4 = dy/dt

Check that (π/6, π/6) is a valid point:
4(π/6)cos(π/6) - (π)tan(π/6) = 0
6.587 = 0
The point is not valid so this question cannot be answered.

-4x*sin(y)(dy/dt) + 4cos(y)(dx/dt) - πsec²(x)(dx/dt) = 0

-4π/6*sin(π/6)(dy/dt) + 4cos(π/6)(1) - πsec²(π/6)(1) = 0

dy/dt = (4π/3 - 2√3)/(-π/3)