Does the density $g(y) \propto (1-y^2)^{(n-3)/2} e^{\delta y} \quad\text{for}\quad |y| \leqslant 1$ have a name? The following probability density function has a particularly simple form, and it was produced when deriving a confidence interval for $\frac{\mu}{\sigma^2}$ , $$g(y;\delta)=c_\delta(1-y^2)^{(n-3)/2}e^{\delta y}\quad\quad \text{for}\quad-1\leq y\leq 1 $$
where $$c_\delta ={{(|\delta |/2)^{n/2-1}}\over{\sqrt{\pi}\, \Gamma((n-1)/2)I_{n/2-1}(|\delta|)}} $$ and where $I_{n/2-1}(|\delta|)$is the modified Bessel function of the first kind of order $n/2-1$.  For a description of the context and derivation of the $g(y;\delta)$ density from the Normal distribution, see Bornholt (2022) (pp. 26, 37).
My question is, does this probability density function have a name already? If not, has anyone seen it used in other settings?
 A: I don't know a name for this distribution, but here is some information that might (or might not) be useful
I have not seen this density function before; a quick search from me did not yield a name.  However, it looks to me like this density might be something that emerges from some kind of monotonic density transform, so I'm going to explore that to see if anything useful comes up.  (You haven't specified your parameter ranges, so I'm going to assume that $\delta>0$
.)  Taking $a \equiv (n-3)/2$ for purposes of simplification, your density function is proportionate to the kernel:
$$g(y) \propto (1-y^2)^a \exp(\delta y) \cdot \mathbb{I}(-1 \leqslant y \leqslant 1).$$
Suppose we now investigate this density only over its support so that we can omit the indicator term.  We will try to find a transformation that is monotonic over this support and gives some simplified form.  Below I will show two transformations that are interesting (hat tip to JimB in comments for the second one).

Sinusoidal transformation: Taking $y = \sin(x)$ we have $dy = \cos(x) \ dx$ which then gives:
$$\begin{align}
g(y) \ dy 
&\propto (1-\sin^2(x))^a \exp(\delta \sin(x)) \ dy \\[12pt]
&= \cos^2(x)^a \exp(\delta \sin(x)) \ dy \\[12pt]
&= \cos^{2a}(x) \exp(\delta \sin(x)) \ dy \\[12pt]
&= \cos^{n-3}(x) \exp(\delta \sin(x)) \ dy \\[8pt]
&= \cos^{n-2}(x) \exp(\delta \sin(x)) \ \frac{dy}{\cos(x)} \\[6pt]
&= \cos^{n-2}(x) \exp(\delta \sin(x)) \ dx. \\[6pt]
\end{align}$$
Consequently, the related random variable $X = \arcsin(Y)$ has the density kernel:
$$f(x) \propto \cos^{n-2}(x) \exp(\delta \sin(x)) \cdot \mathbb{I}(-\tfrac{\pi}{2} \leqslant x \leqslant \tfrac{\pi}{2}).$$
This is also not a density I recognise as any named class, but it is quite an interesting density.  In the special case where $n=3$ this latter density has a closed form for its CDF, given by:
$$F(x) = \frac{\exp(\delta \sin(x)) - \exp(-\delta)}{\exp(\delta) - \exp(-\delta)}
\quad \quad \quad \quad \quad \text{for } -\tfrac{\pi}{2} \leqslant x \leqslant \tfrac{\pi}{2}.$$
For higher values of $n$ the distribution becomes more complicated and it does not have a closed form CDF.

Hyperbolic-tangent transformation: Taking $y = \tanh(x)$ we have $dy = \text{sech}^2(x) \ dx$ which then gives:
$$\begin{align}
g(y) \ dy 
&\propto (1-\tanh^2(x))^a \exp(\delta \tanh(x)) \ dy \\[12pt]
&= \text{sech}^2(x)^a \exp(\delta \tanh(x)) \ dy \\[12pt]
&= \text{sech}^{2a}(x) \exp(\delta \tanh(x)) \ dy \\[12pt]
&= \text{sech}^{n-3}(x) \exp(\delta \tanh(x)) \ dy \\[8pt]
&= \text{sech}^{n-1}(x) \exp(\delta \tanh(x)) \ \frac{dy}{\text{sech}^2(x)} \\[6pt]
&= \text{sech}^{n-1}(x) \exp(\delta \tanh(x)) \ dx \\[6pt]
\end{align}$$
Consequently, the related random variable $X = \text{arctanh}(Y)$ has the density kernel:
$$f(x) \propto \text{sech}^{n-1}(x) \exp(\delta \tanh(x)).$$
This is also not a density I recognise as any named class, but it is quite an interesting density.  In the special case where $n=3$ this latter density has a closed form for its CDF, given by:
$$F(x) = \frac{\exp(\delta \tanh(x)) - \exp(-\delta)}{\exp(\delta) - \exp(-\delta)}
\quad \quad \quad \quad \quad \text{for } x \in \mathbb{R}.$$
For higher values of $n$ the distribution becomes more complicated and it does not have a closed form CDF.

Hypergeometric transformation: We can abstract from the above transformations to use a transformation $y = H(x)$ satisfying the nonlinear ordinary differential equation:
$$H'(x) = (1-H(x)^2)^{-a}.$$
The solution to this differential equation uses a hypergeometric function:
$$x = \text{const} +  y \cdot {_2}F_1(\tfrac{1}{2}, -a; \tfrac{3}{2}; y).$$
It lets us write:
$$\begin{align}
g(y) \ dy 
&\propto (1-H(x)^2)^a \exp(\delta H(x)) \ dy \\[12pt]
&= \frac{1}{H'(x)} \exp(\delta H(x)) \ dy \\[12pt]
&= \exp(\delta H(x)) \ dx, \\[12pt]
\end{align}$$
which gives:
$$f(x) \propto \exp(\delta H(x)) \cdot \mathbb{I}(x \in \mathscr{X}).$$
This is a complicated distribution, since it involves the hypergeometric function.  It is also not a named class of distributions that I'm aware of.
