$$\int\limits_0^\infty \exp\Big(-x^2-\frac{a^2}{x^2}\Big)\,\mathrm dx = \frac{\sqrt{\pi}}2e^{-2|a|}$$
How do we prove the above result? I tried using the Leibniz Integral rule but I ended up getting an indefinite answer because of the $x^2$ in the denominator in the power.
You have only requested for a solution using differentiation under the integral sign, but I've two solutions. I'll be sharing both of them.
Using differentiation under the integral sign:
For $a\ge 0$, let $$I(a) = \int\limits_0^\infty\exp\Big(-x^2-\frac{a^2}{x^2}\Big)\,\mathrm dx$$
Differentiating both sides w.r.t. $a$, we get $$I'(a) = -2\int\limits_0^\infty\frac{a}{x^2}\exp\Big(-x^2-\frac{a^2}{x^2}\Big) \,\mathrm dx$$
Substituting $x\mapsto\frac ax$, we get
$$I'(a) = -2 \int\limits_0^\infty\exp\Big(-x^2-\frac{a^2}{x^2}\Big)\,\mathrm dx = -2I(a) $$ The general solution for this differential equation will be $I(a) =k_1 e^{-2a}$. Noting that $I(0) = \frac{\sqrt\pi}2$, we get $k_1 = \frac{\sqrt\pi}2$. Thus, for $a\geq0$ $$ \int\limits_0^\infty\exp\Big(-x^2-\frac{a^2}{x^2}\Big)\,\mathrm dx = \frac{\sqrt\pi}2e^{-2a}$$ Putting modulus around $a$ extends the result $\forall a\in \mathbb R$.
Using substitution:
In the same $I(a) $, substitute $x\mapsto \frac ax$. This gives
$$I(a) = \int \limits_0^\infty\frac a{x^2} \exp\Big(-x^2-\frac {a^2}{x^2}\Big) \,\mathrm dx$$
Adding the two equations,
$$I(a) = \frac12 \int\limits_0^\infty\exp\Big(-x^2-\frac {a^2}{x^2}\Big)\Big(1+\frac a{x^2}\Big)\,\mathrm dx $$ Now, we note that $x^2+\frac{a^2}{x^2}= (x-\frac ax)^2+2a $. Keeping this in mind, we substitute $t=x-\frac ax$. This gives
$$I(a) = \frac{e^{-2a}}2 \int\limits_{-\infty} ^\infty e^{-t^2}\,\mathrm dt $$ The later integral is the Gaussian integral, equal to $\sqrt\pi$. After using the fact that the integrand is even, we had used this result to calculate $k_1$. Using this result, we get the same result as before.
Comment if any mistakes or confusion regarding the steps.