Let $D$ be an open subset of $\mathbb{R}^3$, and $f: D \to \mathbb{R}$ be a smooth function whose gradient $ \nabla f \neq 0$ on $D$. Consider the surface $M = \{(x_1,x_2,x_3) \in D \mid f(x_1,x_2,x_3)=0 \} $. Assign at each $p \in M$ the unit normal vector $N(p)$ from $\nabla f$, and the Riemannian metric $g$ induced on $M$ from the usual flat metric on $\mathbb{R}^3$.
I'm not sure what the induced metric means. Is it the same metric or something different? I tried to search for some answer but could not find any.
Since $f$ is smooth and $\nabla f \neq 0$, $M$ is a smooth submanifold of $\Bbb R^3$, and at each point $p \in M$, we can identify $$T_p M = \ker (df)_p = \langle (\nabla f)_p \rangle^{\perp} = \langle N(p) \rangle^{\perp} \subset T_p \Bbb R^3$$ Then, the induced metric (or pullback metric) on $M$ is the pullback $\iota^* \bar g$, where $\bar g$ is the flat metric on $\Bbb R^3$: For any $p \in M$ and $X, Y \in T_p M$, $$(\iota^* \bar g)_p(X, Y) = \bar g_p(T_p \iota \cdot X, T_p \iota \cdot Y) .$$ Under the above identification of $T_p M$ with $T_p \Bbb R^3$, we're simply regarding vectors in $T_p M$ as tangent vectors in $T_p \Bbb R^3$, and so we may as well write this quantity as $$(\iota^* \bar g)_p(X, Y) = \bar g_p(X, Y).$$ Despite this formula, the metrics are emphatically not the same: $\bar g$ and the induced metric $\iota^* \bar g$ live on different manifolds, and $\iota^* \bar g$ contains strictly less information than $\bar g$.