Explaining Tangent Planes With Implicit Differentiation for Multivariable Calculus Let $f : D \subseteq\Bbb R^3 \to \Bbb R$ be a $C^1$ function. A point $(a, b, c) \in \Bbb R^3$ is a critical point of $f$ if 
$$
\nabla f(a, b, c) = (0, 0, 0).
$$ 
The corresponding number $k = f(a, b, c)$ is called a critical value: any real number $c$ that is not a critical value is called a regular value. (Note that if $c$ is a regular value, then this means there are no critical points of $f$ in the level set defined by $f(x, y, z) = c$.) 
Use implicit differentiation to explain the following fact: Let $c$ be a regular value of $f$. If $S \subset\Bbb R^3$ is the surface defined by $S = \{(x, y, z)| f(x, y, z) = c\}$, then at every point $(a, b, c) \in S$, the surface $S$ can be approximated by a tangent plane with normal vector given by $\nabla f(a, b, c)$. 
Your argument should be along the following lines: we know how to approximate a graph by a tangent plane. If the partial derivative of $f$ with respect to some variable is non-zero at a point, then near that point, $S$ is given by a graph of the form... (This last bit is a consequence of the Implicit Function Theorem that you can accept without proof. If you want to see the precise statement, it’s in the Week 3 handout on the proof of the chain rule.)
 A: Take the equation defining the surface $S$: $f(x,y,z) = c$. Suppose that locally, we can solve for $z$ in this equation. For example, if the equation was $x^2+y^2+z^2=c$, we could solve $z = \sqrt{c-x^2-y^2}$. Then we may assume $z = z(x,y)$ is a function of $x$ and $y$ (this is what the comment about the Implicit Function Theorem is referring to). Maybe you can't always solve for $z$, and you have to solve for $x$ or $y$ instead, but the idea is the same...
You probably know that if you have a graph $z = z(x,y)$, then the normal vector to the tangent plane is $\left(-z_x, \, -z_y, \, 1 \right)$. Here I'm using subscripts to denote partial derivatives, so $z_x = \frac{\partial z}{\partial x}$.
The equation above can be written now as $f(x,y,z(x,y)) = c$. Use the chain rule to differentiate this equation and solve for $z_x$ and $z_y$. First let's do with respect to $x$:
$$ f_x + f_z z_x = 0 $$
This tells you that $z_x = \frac{-f_x}{f_z}$. You can do a similar calculation with $y$. So, with the comment above, the normal vector to $S$ is given by
$$ \left(\frac{f_x}{f_z}, \, \frac{f_y}{f_z}, \, 1 \right) $$
If you rescale a normal vector, it is still a normal vector, so we could also rescale by $f_z$ and take $(f_x,f_y,f_z)$ instead. This is exactly the gradient $\nabla f$.
