= Solution
Put $h(t,x)=e^{-x^2-t^2/x^2}$. For every $t\geq0$, $0<h(t,x)\leq e^{-x^2}$, and the upper bound is a <Lebesgue integrable function> by part (b)(i). Since $h(t,x)\to e^{-x^2}$ for each $x>0$ as $t\downarrow0$, the <dominated convergence theorem> gives right <continuity> at zero and
$$
F(0)=\lim_{t\downarrow0}F(t)=\frac{\sqrt\pi}{2}.
$$
To apply <dominated differentiation under the integral sign>, fix $t_0>0$ and restrict $t$ to $[t_0/2,3t_0/2]$. Then
$$
|\partial_t h(t,x)|=\frac{2t}{x^2}e^{-x^2-t^2/x^2}
\leq D(x):=3t_0x^{-2}e^{-x^2-t_0^2/(4x^2)}.
$$
For $x\geq1$, $D(x)\leq3t_0e^{-x^2}$. For $0<x<1$, writing $a=t_0^2/4>0$, the <substitution rule> $u=1/x$ gives
$$
\int_0^1x^{-2}e^{-a/x^2}\,dx=\int_1^\infty e^{-au^2}\,du<\infty.
$$
Thus $D$ is integrable. The <mean value theorem> bounds nearby difference quotients by $D$, so the <dominated convergence theorem> justifies the derivative at $t_0$. As $t_0$ was arbitrary,
$$
F'(t)=-2t\int_0^\infty x^{-2}e^{-x^2-t^2/x^2}\,dx,\qquad t>0.
$$
Now make the <change of variables> $u=t/x$. The reversal of limits cancels the negative differential and gives
$$
\int_0^\infty x^{-2}e^{-x^2-t^2/x^2}\,dx
=\frac1t\int_0^\infty e^{-u^2-t^2/u^2}\,du=\frac{F(t)}t.
$$
Consequently $F'(t)=-2F(t)$. This also follows in sign from the fact that the positive integrand decreases strictly as $t$ increases: the printed equation with a positive sign cannot hold. Solving the <linear differential equation>, or differentiating $e^{2t}F(t)$, and using right <continuity> at zero gives
$$
\boxed{F'(t)=-2F(t)\ (t>0),\qquad F(t)=\frac{\sqrt\pi}{2}e^{-2t}\ (t\geq0).}
$$
This is a half-line instance of the <Gaussian integral with an inverse-square term>.
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