Let $f(x,z)$ be convex in $(x,z)$ and define $g(x)=\inf_zf(x,z)$. Express the conjugate $g^*$ in terms of $f^*$.
Solution: The (convex) conjugate of $f(x,z)$ is: $$f^*(y,s)=\sup_{(x,z)\in dom~f}\left(y^Tx+sz-f(x,z) \right)$$
Also, the conjugate of $g(x)$ is: \begin{align} g^*(y)&=\sup_{x\in dom~g} \left(y^Tx-g(x)\right)\\ &=\sup_{x\in dom~g} \left(y^Tx-\inf_zf(x,z)\right) \end{align} Can I proceed as follows?
let $z^*$ be a global minimizer for the convex function $f(x,z)$, then \begin{align} g^*(y)&=\sup_{x\in dom~g} \left(y^Tx-f(x,z^*)\right)\\ &=\sup_{x\in dom~g} \left(y^Tx-f(x,z^*)\right)+sz^*-sz^*\\ &=f^*(y,z*)-sz^* \end{align}
Your bivariate function has two "pictures" attached to it, in that if you define for every pair $(x, y)$ auxiliary functions $v_z := f(., z)$ and $u_x := f(x, .)$, then it holds that $u_x(z) = u_z(x)$ for all pairs $(x, y)$. Now, define the maginal: $g(x) := \inf_z u_x(z)$, for ever $x$. One computes readily
$$ \begin{split} g^*(y) &:= \sup_x \langle x, y\rangle - g(x) = \sup_x \langle x, y\rangle - \inf_z u_x(z)\\ &= \inf_z \sup_x \langle x, y\rangle - u_x(z) = \inf_z \sup_x \langle x, y\rangle - v_z(x) = \inf_z v_z^*(y). \end{split} $$
In plain english, we've shown that $$(x \mapsto \inf_z f(x, z))^*(y) = \inf_z (x \mapsto f(x, z))^*(y),\; \forall y$$