\left(\frac{\cos 3x}{\cos 4x}\right)^{\frac 1 {x^2}}=\left(1+\frac{\cos 3x}{\cos 4x}-1\right)^{\frac 1 {x^2}}=\left(1+\frac{\cos 3x- \cos 4x}{\cos 4x}\right)^{\frac 1 {x^2}}=\left[\left(1+\frac{\cos 3x- \cos 4x}{\cos 4x}\right)^{\frac {\cos 4x}{\cos 3x - \cos 4x}}\right]^{\frac {\cos 3x - \cos 4x}{x^2 \cos 4x}}