= Solution
Expand the exponent in the <Inverse Gaussian distribution> with unit <shape parameter>:
$$
-\frac{(x-\mu)^2}{2\mu^2x}
=-\frac{x}{2\mu^2}+\frac1\mu-\frac1{2x}.
$$
Consequently its <probability density function> can be written as
$$
f(x;\mu)=h(x)\exp\{\eta x-A(\eta)\},
$$
where
$$
h(x)=\frac1{\sqrt{2\pi x^3}}\exp\left(-\frac1{2x}\right),
\qquad
\eta=-\frac1{2\mu^2}<0,
\qquad
A(\eta)=-\sqrt{-2\eta}.
$$
Indeed, $-A(\eta)=\sqrt{-2\eta}=1/\mu$. This is the canonical form of a one-parameter <exponential family>, and its <natural parameter of an exponential family>[natural parameter] is therefore
$$
\boxed{\eta=-\frac1{2\mu^2}}.
$$
Here the natural statistic is $T(X)=X$. The <exponential-family derivative identities> give
$$
\mathbb E_\eta X=A'(\eta)=(-2\eta)^{-1/2}=\mu
$$
and
$$
\operatorname{var}_\eta X=A''(\eta)=(-2\eta)^{-3/2}=\mu^3.
$$
Thus
$$
\boxed{\mathbb E X=\mu,
\qquad \operatorname{var}X=\mu^3}.
$$
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