The answer is no.
Theorem. Let $f:\mathbb{R}^d\to\mathbb{R}$ be any continuous function. Then there is a Lipschitz function $F:\mathbb{R}^d\times\mathbb{R}\to\mathbb{R}$ such that for every $x\in\mathbb{R}^d$, $F(x,y)=0$ if and only if $y=f(x)$.
Proof. Let $G\subset\mathbb{R}^{d}\times\mathbb{R}$ be the graph of $f$. Then the set $G$ is closed, and the function $F:\mathbb{R}^d\times\mathbb{R}\to\mathbb{R}$ defined by
$$
F(x,y)=\operatorname{dist}((x,y),G)
$$
is $1$-Lipschitz continuous. Since $F$ vanishes exactly on $G$,we have that for every $x\in\mathbb{R}^d$, $F(x,y)=0$ if and only if $(x,y)\in G$, i.e. $y=f(x)$. $\Box$
There are examples of Lipschitz functions $F(x,y)$ as in your problem, with the zero set $y=f(x)$ being given by a discontinuous function $f$.
Example. Let $f:\mathbb{R}\to\mathbb{R}$, $f(x)=1/x$ if $x\neq 0$, and $f(0)=0$. Note that the graph $G$ of $f$ is a closed subset of $\mathbb{R}^2$. Therefore, the above construction gives $F(x,y):\mathbb{R}^2\to\mathbb{R}$ with the zero zet being the graph of $y=f(x)$ which is a discontinuous function. Note however, that the function $y=f(x)$ is unbounded.
$\Box$
Theorem. Suppose that $F:\mathbb{R}^n\times\mathbb{R}^m\to\mathbb{R}$ is continuous and $M>0$ is a constant. If for every $x\in\mathbb{R}^n$ there is a unique $y=y(x)\in\mathbb{R}^m$ such that
$$
|y(x)|\leq M
\quad
\text{and}
\quad
F(x,y(x))=0,
$$
then the function $\mathbb{R}^n\ni x\mapsto y(x)\in\mathbb{R}^m$ is continuous.
Proof. Suppose that $x_i\to x$. We need to prove that $y(x_i)\to y(x)$. Suppose to the contrary that this is not the case so there is $\varepsilon$ and a subsequence such that
$$
(*)
\quad \quad \quad
|y(x_{i_j})-y(x)|\geq\varepsilon.
$$
Since $|y(x_{i_j})|\leq M$ is bounded, we can select a convergent subsequence $y(x_{i_{j_k}})\to y$, $|y|\leq M$. Continuity of $F$ yields
$$
0=F(x_{i_{j_k}},y(x_{i_{j_k}}))\to F(x,y).
$$
That is $|y|\leq M$ and $F(x,y)=0$, which by the assumptions in the theorem mean that $y=y(x)$, so $y(x_{i_{j_k}})\to y(x)$ and we arrive at a contradiction with (*). $\Box$
