Should be this on the fourth line:
= π + π sin(πx) 2cos(π x)/cos^{4}(π x)
i.e. Missed a π inside the sin(x).

And the final answer:
f'(x) = π + 2π sec^{2}(π x)tan(πx)
f'(1/4) ≈ 15.708

For your second question, you surprisingly did get the correct value for f'(-1), but the function you found is in general wrong:
f(x) = [(1/x^{2}-2) / (1/x^{2})]^{2}
f(x) = [(1/x^{2}-2) *x^{2}]^{2}
f(x) = (1/x^{2}-2)^{2}*x^{22}
f(x) = (1/x^{4}-4/x^{2}+4)x^{4}
f(x) = 1-4x^{2}+4x^{4}
(as Tom derived)

From this it far easier to find f'(x):
f'(x)= -8x+16x^{3}
f'(x) = 8x(2x^{2}-1)
f'(-1) = -8
This is similar to be not the same as the function you derived.