Derive DTFT of $x[2n]$
HINT:

If you have a sequence

$$\hat{x}[n]=\begin{cases}x[n],&amp;n\text{ even}\\0,&amp;n\text{ odd}\end{cases}$$

then the DTFT of $x[2n]$ can be written as

$$\sum_{n=-\infty}^{\infty}x[2n]e^{-jn\omega}=\sum_{n=-\infty}^{\infty}\hat{x}[n]e^{-jn\omega/2}=\hat{X}(e^{j\omega/2})\tag{1}$$

You can obtain $\hat{x}[n]$ as

$$\hat{x}[n]=\frac12\left(x[n]+(-1)^nx[n]\right)\tag{2}$$

Taking the DTFT of $(2)$ to express $\hat{X}(e^{j\omega})$ in terms of $X(e^{j\omega})$, and using $(1)$ will give the desired result.