find the point of (x^2+4)y=8 which has a horizontal tangent line (use implicit to find dy/dx)?

Differentiating each side wrt x [recalling y = y(x)] gives :

2x * y + (x² + 4)*y' = 0.

A horizontal tangent is when y'=0. Put this condition in the above eqs. Conclusions?

　
(x²+4)y = 8

Differentiate implicitly:

(2x+0)y + (x²+4) dy/dx = 0
(x²+4) dy/dx = −2xy
dy/dx = −2xy/(x²+4)

Curve has horizontal tangent when dy/dx = 0
−2xy/(x²+4) = 0
x = 0 or y = 0

When y = 0 we get:
(x²+4)0 = 8
0 = 8
Not possible

When x = 0 we get:
(0²+4)y = 8
y = 2

Horizontal tangent at point (0,2)

We have y = 8/(x^2+4)
dy/dx = -16x/(x^2+4)^2
dy/dx = 0 if x = 0; hence, y = 8/4=2
(0,2) is such point.