Clearly, this problem is meant to be solved using Fourier Transform methods.
I'm using the definition that the Fourier transform of $f(t)$ is $F(\omega) = \displaystyle\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt$.
Accordingly, the inverse Fourier transform of $F(\omega)$ is $f(t) = \dfrac{1}{2\pi}\displaystyle\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega$.
Define $g(z) = \log \left( \displaystyle\int_{-\infty}^\infty e^{-(-z-t)^2} f(t) dt\right)$.
Then, we have $\displaystyle\int_{-\infty}^\infty e^{-y^2}g(x-y)\,dy=-cx^2$.
The left hand side is simply the convolution of $e^{-x^2}$ and $g(x)$.
The Fourier transform of $e^{-x^2}$ is $\sqrt{\pi}e^{-\omega^2/4}$ and the Fourier transform of $-cx^2$ is $2\pi c\delta''(\omega)$.
By the convolution property, we get $\sqrt{\pi}e^{-\omega^2/4}G(\omega) = 2\pi c\delta''(\omega)$, i.e. $G(\omega) = 2c\sqrt{\pi}e^{\omega^2/4}\delta''(\omega)$.
From Wikipedia, the Dirac Delta function satisfies $x\delta'(x) = -\delta(x)$. Differentiation yields $x\delta''(x)+\delta'(x) = -\delta'(x)$, i.e. $x\delta''(x) = -2\delta'(x)$. Therefore, $x^2\delta''(x) = -2x\delta'(x) = 2\delta(x)$. Then, since $x^n\delta(x) = 0$ for integers $n \ge 1$, we have $x^n\delta''(x) = 0$ for integers $n \ge 3$.
Using Taylor expansion on $e^{\omega^2/4}$, we get $G(\omega) = 2c\sqrt{\pi}e^{\omega^2/4}\delta''(\omega) = c\sqrt{\pi}(2\delta''(\omega)+\delta(\omega))$.
Taking the inverse Fourier transform yields $g(z) = -\dfrac{c}{\sqrt{\pi}}\left(z^2-\dfrac{1}{2}\right)$.
Now, we've reduced the problem to solving $\displaystyle\int_{-\infty}^\infty e^{-(-z-t)^2} f(t)\,dt = e^{-\tfrac{c}{\sqrt{\pi}}\left(z^2-\tfrac{1}{2}\right)}$.
Using the substitution $z \to -z$, we get $\displaystyle\int_{-\infty}^\infty e^{-(z-t)^2} f(t)\,dt = e^{\tfrac{c}{2\sqrt{\pi}}}e^{-\tfrac{c}{\sqrt{\pi}}z^2}$.
The left hand side is simply the convolution of $e^{-z^2}$ and $f(z)$.
The Fourier transform of $e^{-z^2}$ is $\sqrt{\pi}e^{-\tfrac{\omega^2}{4}}$ and the Fourier transform of $e^{\tfrac{c}{2\sqrt{\pi}}}e^{-\tfrac{c}{\sqrt{\pi}}z^2}$ is $e^{\tfrac{c}{2\sqrt{\pi}}}\sqrt{\tfrac{\pi\sqrt{\pi}}{c}}e^{-\tfrac{\sqrt{\pi}}{c}\tfrac{\omega^2}{4}}$
By the convolution property, we get $\sqrt{\pi}e^{-\tfrac{\omega^2}{4}}F(\omega) = e^{\tfrac{c}{2\sqrt{\pi}}}\sqrt{\tfrac{\pi\sqrt{\pi}}{c}}e^{-\tfrac{\sqrt{\pi}}{c}\tfrac{\omega^2}{4}}$.
Now, solve for $F(\omega)$, and take the inverse Fourier transform to get the answer.