Sorry, but I'm afraid your assumption is wrong, and Wolfram is right.
All the limits are taken as x -> 0.
Lim ((cos(x)/cos(2x))^(1/x^2)
= Lim exp((1/x^2)*ln(cos(x)/cos(2x)))
= exp(Lim (1/x^2)*ln(cos(x)/cos(2x)))
= exp(Lim (1/x^2)*(ln(cos(x)) – ln(cos(2x)))
= exp(Lim (1/x^2)*ln(cos(x)) – (1/x^2)*ln(cos(2x)))
Using l’Hôpital’s rule on Lim (1/x^2)*ln(cos(x)) and Lim (1/x^2)*ln(cos(2x)), we get
Lim (1/x^2)*ln(cos(x)) = -sin(x)/(2x.cos(x))
Lim (1/x^2)*ln(cos(2x)) = -2sin(2x)/(2x.cos(2x))
From which we continue with
= exp(Lim (1/2)(-sin(x).cos(2x) + 2sin(2x).cos(x)) / (x.cos(2x).cos(x)))
= exp((1/2)*Lim (-sin(x).cos(2x) / (x.cos(2x).cos(x)) + 2*Lim(sin(2x).cos(x) / (x.cos(2x).cos(x)))
= exp((1/2)*Lim (-sin(x) / (x.cos(x)) + 2*Lim(sin(2x) / (x.cos(2x)))
= exp((1/2)*[Lim (-sin(x) / (x)*Lim(1/cos(x)) + 2*Lim(2sin(x).cos(x) / (x.cos(2x))])
= exp((1/2)*[(-1)*(1) + 2*Lim(2sin(x)/x)*Lim(cos(x) / (2cos^2(x) – 1)])
= exp((1/2)*[-1 + 2*2*(1) / (2 – 1)])
= exp((1/2)*[-1 + 4])
= exp(3/2)