Differentiate both sides of
2 + 2 sin(x) = 3 tan(y)
with respect to a new variable t, assuming both x and t depend on t, which will involve the chain rule:
d(2 + 2 sin(x))/dt = d(3 tan(y))/dt
d(2)/dt + d(2 sin(x))/dt = 3 d(tan(y))/dt
0 + 2 d(sin(x))/dt = 3 sec²(y) dy/dt
2 cos(x) dx/dt = 3 sec²(y) dy/dt
Now, when x = π/6, y = π/4, and dx/dt = 2, solve for dy/dt :
2 cos(π/6) × 2 = 3 sec²(π/4) dy/dt
dy/dt = 4 cos(π/6) / (3 sec²(π/4))
dy/dt = 4 (√3/2) / (3 (√2)²)
dy/dt = √3/3 = 1/√3