The correct answer is $\boxed{-\frac{1}{2}X\left(\frac{f}{2}\right)e^{-j2\pi f}}$.
The Fourier transform of a function $x(t)$ is defined as
$$X(f) = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt$$
The inverse Fourier transform is defined as
$$x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(f) e^{j2\pi ft} df$$
Given that $x(t)$ is an even function, its Fourier transform is given by
$$X(f) = \frac{1}{2} \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt = \frac{1}{2} \int_{-\infty}^{\infty} x(-t) e^{j2\pi ft} dt = \frac{1}{2} X(-f)$$
The Fourier transform of $y(t)$ is given by
$$Y(f) = \int_{-\infty}^{\infty} y(t) e^{-j2\pi ft} dt = \int_{-\infty}^{\infty} -x(t-\tau) e^{-j2\pi ft} dt = -X(f) e^{-j2\pi \tau f}$$
where $\tau$ is the delay parameter.
In this case, $\tau = \frac{1}{2}$, so
$$Y(f) = -X(f) e^{-j2\pi \frac{1}{2} f} = -\frac{1}{2}X\left(\frac{f}{2}\right)e^{-j2\pi f}$$