I have this statistical model:
$$f_{j,k}(y)=\frac{\sqrt{j}}{\sqrt{2 \pi}}e^{\sqrt{jk}}y^{-\frac{1}{2}} \text{exp}\left( -\frac{1}{2} (j y + \frac{k}{y}) \right) \quad \quad y>0$$
In this case the parameters j and k are unknown.
My question is, why is this model a part of the exponential family?
A family of distributions is said to belong to a vector exponential family if the probability density function can be written as $$ f_{\pmb{\theta}}(x) =h(x) \exp\left(\sum_{i=1}^k \eta_i(\pmb{\theta}) T_i(x) - A(\pmb{\theta}) \right)= h(x) \exp\left(\pmb{\eta}(\pmb{\theta})\cdot \pmb{T}(x) - A(\pmb{\theta}) \right) $$ or in the alternative, equivalent form $$ f_{\pmb{\theta}}(x) =h(x)g(\pmb{\theta}) \exp\left(\sum_{i=1}^k \eta_i(\pmb{\theta}) T_i(x) \right)= h(x)g(\pmb{\theta}) \exp\left(\pmb{\eta}(\pmb{\theta})\cdot \pmb{T}(x)\right) $$ The exponential family is said to be in canonical form if $\eta_i(\pmb{\theta})=\theta_i$, for all $i$.
In your case $$ f_{j,k}(y)=\frac{\sqrt{j}}{\sqrt{2 \pi}}e^{\sqrt{jk}}y^{-\frac{1}{2}} \text{exp}\left( -\frac{1}{2} (j y + \frac{k}{y}) \right) \quad \quad y>0 $$ that is an exponential family in canonical form with $$ \begin{align} \pmb{\theta}&=(j,k)\\ h(y)&=\frac{1}{\sqrt{2 \pi}}y^{-\frac{1}{2}}\\ g(\pmb{\theta})&=g(j,k)=\sqrt{j}\,\operatorname{e}^{\sqrt{jk}}=\operatorname{e}^{-A(\pmb\theta)}\\ \pmb{\eta}(\pmb{\theta})&=\left(\eta_1(j,k),\eta_2(j,k)\right)=(j,k)=\pmb{\pmb\theta}\\ \pmb T(y)&=\left(T_1(y),T_2(y)\right)=\left(-\frac{y}{2},-\frac{1}{2y}\right) \end{align} $$