How do I find the complex Fourier transform of $f(x) = \dfrac{1}{1 + x^2 + x^4}$?
I know that the complex Fourier transform is $\hat{f}(k) = \displaystyle \int_{-\infty}^{\infty} f(x)e^{-ikx}\ dx$.
Another trick I thought of doing is writing $x^2 + x^4 + 1$ as $(x^2 - x + 1)(x^2 + x + 1)$ where the roots of $x^2 - x + 1$ are $x = \dfrac{1 \pm \sqrt{3}i}{2}$ and the roots of $x^2 + x + 1$ are $x = \dfrac{-1 \pm \sqrt{3}i}{2}$. But then how do I calculate the Residue to find the complex Fourier transform.
Also, I need to use direct integration to find the inverse Fourier transform of the $\hat{f}(k)$ that I find and check that I recover $f(x) = \dfrac{1}{1 + x^2 + x^4}$
Let me show you what I think is the easiest way to compute the residues. You've already shown that
$$ f(z) = \frac{1}{(z - \alpha_1)(z - \alpha_2)(z - \alpha_3)(z - \alpha_4)},$$
where $\alpha_1, \alpha_2, \alpha_3, \alpha_4$ are the four roots that you computed in your original question.
So how do we determine the residue at $z = \alpha_1$ (say)?
Let's consider the function
$$ g(z) = \frac{1}{(z - \alpha_2)(z - \alpha_3)(z - \alpha_4)},$$ defined on $\mathbb C \setminus \{ \alpha_2, \alpha_3, \alpha_4 \}$.
If $B_\delta(\alpha_1)$ is an open disk around $\alpha_1$, small enough that it doesn't contain $\alpha_2, \alpha_3, \alpha_4$, then $g$ is holomorphic on $B_\delta(\alpha_1)$. So $g$ can be represented by a Taylor series on $B_\delta(\alpha_1)$: $$ g(z) = g(\alpha_1) + g'(\alpha_1) (z - \alpha_1) + \frac 1 {2!} g''(\alpha_1) (z - \alpha_1)^2 + \dots$$
On the punctured open disk $B_\delta ' (\alpha_1) = B_\delta (\alpha_1) \setminus \{ \alpha_1 \}$, we have $g(z) = (z - \alpha_1) f(z)$. Thus, on this punctured disk, we have $$ f(z) = \frac{g(\alpha_1)}{z - \alpha_1} + g'(\alpha_1) + \frac 1 {2!} g''(\alpha_1) (z - \alpha_1) + \dots$$ This is the Laurent series for $f(z)$! So the coefficient of the $(z - \alpha_1)^{-1}$ term in this series the residue of $f$ at $\alpha_1$. Thus $$ \text{Res}_{z = \alpha_1} = g(\alpha_1). $$
It remains to evaluate $g(\alpha_1)$. That's easy - it's $$ g(\alpha_1) = \frac{1} {(\alpha_1 - \alpha_2)(\alpha_1 - \alpha_3)(\alpha_1 - \alpha_4)}$$
Taking a step back, it's worth noting that this method can be used to compute the residue of any meromorphic function at a simple pole. If $f(z)$ has a simple pole at $\alpha$, then $g(z) = (z - \alpha)f(z)$ has a removable singularity at $\alpha$ and $\lim_{z \to \alpha} g(z)$ is the constant term in the Taylor series for $g$ at $\alpha$, which is also the coefficient of the $(z - \alpha)^{-1}$ term in the Laurent series for $f$ at $\alpha$ (i.e. the residue of $f$ at $\alpha$).