Let $g(x,y)\in C^1(\mathbb{R}^n\times\mathbb{R}^n,\mathbb{R})$ such that the least eigenvalue of $\nabla_{yy} g(x,y)$ is always greater than $1$, $\nabla_{yy} g(x,y)$ is invertible and $\nabla_{yy} g(x,y)$ is bounded.
The function $G(x,y,z):=z\cdot y +g(x,y)$ must attain a minimum at the point $y_{x,z}$ for each fixed $x,z$, as $G$ is convex in $y$ ($\nabla_{yy} G(x,y,z)=\nabla_{yy} g(x,y)$ is positive definite). Hence, we have $\nabla_y G(x,y_{x,z},z)=z+\nabla_y g(x,y_{x,z})=0$ for any $x,z$.
Let $(x_0,z_0) \in \mathbb{R}^n \times \mathbb{R}^n$, there is a point $y_{x_0,z_0}$ such that $\nabla_y G(x_0,y_{x_0,z_0},z_0)=0$ by the above arguments. Using the implicit function theorem, there is an open set $U_0$ containing the point $(x_0,z_0)$ and a unique $C^1$ function $f_0(x,z)$ defined on $U_0$ such that $\nabla_y G(x,f_0(x,z),z)=0$ on $U_0$ and $f_0(x_0,z_0)=y_{x_0,z_0}$.
My problem is, can such $f$ be defined globally(at least continuously)? I am not sure if my arguments in the following is correct. Assuming the notations $(x_0,z_0)$, $U_0$ and $f_0$ in the above, we let $(x_1',z_1')$ be some point on the boundary of $U_0$. Using the implicit function theorem again, there is an open set $U_1$ containing the point $(x_1',z_1')$ and a unique $C^1$ function $f_1(x,z)$ defined on $U_1$ such that $\nabla_y G(x,f_1(x,z),z)=0$ on $U_1$. There is a pair of rational numbers $(x_1,z_1)$ close enough to $(x_1',z_1')$ and $\delta_1>0$ such that $B_{\delta_1}((x_1,z_1))\subset U_1$ and $B_{\delta_1}((x_1,z_1))\cap U_0$ is non-empty. Then for any point in $B_{\delta_1}((x_1,z_1)) \cap U_0$, we should have $f_0=f_1$ as the implicit function theorem provides the uniqueness. If we carry on this process, can we collect all these countably many balls to conclude that there is a globally defined implicit $f$ (by gluing all these $f_0,f_1,...$)?