Show that both of the turning points of the curve of y=4-x^2-(16/(x)^2) are local max point.

Show that both of the turning points of the curve of y=4-x^2-(16/(x)^2) are local max point. Determine their coordinates. Thanks in advance giving best answer

y=4-x^2-(16/(x)^2)
y ' = - 2x + (32/x^3) = ( - 2x^4 + 32)/x^3
Solve y ' = 0
( - 2x^4 + 32)/x^3 = 0
( - 2x^4 + 32)/x^3 = 0
x^4 - 16 = 0
(x^2 + 4)(x - 2)(x + 2) = 0
real solutions are x = 2 or x = - 2
They (x = ± 2) are max points ( study the sign of y ' = ( - 2x^4 + 32)/x^3 to the left and to the right of x = ± 2 )
Max points are :
( -2, - 4) and (2, - 4 )