Sufficient conditions for Implicit function theorem

Question

I need to determine whether the surface whose equation is $xy - y (logz) +sin(xz) = 0 $ can be represented in the form $z= f(x,y)$ near $(0,2,1)$ using implicit function theorem. Assume $f$ to be a smooth function.

If I apply Implicit function theorem here and calculate the $M(x,y,z)$ matrix formed by differentiating the function $xy - y (logz) +sin(xz) = f(x,y,z)$ with respect to $z$ then at $(0,2,1)$, I am getting this matrix to be invertible. Is this sufficient to show the required condition?

Answer 1

Let's see (from Wikipedia):

Let $f: \Bbb R^{n+m} → \Bbb R^m$ be a continuously differentiable function, and let $\Bbb R^{n+m}$ have coordinates $(\mathbf x, \mathbf y)$. Fix a point $(\mathbf a, \mathbf b) = (a_1, ..., a_n, b_1, ..., b_m)$ with $f(\mathbf a, \mathbf b) = \mathbf c$, where $\mathbf c \in \Bbb R^m$. If the matrix $\left [\frac{\partial f_i}{\partial y_j}(a, b)\right ]$ is invertible, then there exists an open set $U$ containing $\mathbf a$, an open set $V$ containing $\mathbf b$, and a unique continuously differentiable function $g: U \to V$ such that $$\{ (\mathbf{x}, g(\mathbf{x}))|\mathbf x \in U \} = \{ (\mathbf{x}, \mathbf{y}) \in U \times V| f(\mathbf{x}, \mathbf{y}) = \mathbf{c} \}.$$

$f$ is continuously differentiable? Yes. Derivative matrix of $f$ is invertible? Yes.

Looks like you're covered.