The gradient points in the direction of fastest increase, and the rate of change is zero along a level curve. Put those together: the gradient must be perpendicular to the level curves. In three variables the same reasoning makes $\nabla F$ perpendicular to level surfaces, and that gives tangent planes to surfaces that are not graphs.
Key idea
$\nabla F(P)$ is a normal vector to the level surface $F(x, y, z) = k$ at $P$. The tangent plane there is
$$F_x(P)(x - x_0) + F_y(P)(y - y_0) + F_z(P)(z - z_0) = 0,$$
and the normal line is $\mathbf{r}(t) = P + t\,\nabla F(P)$. In two variables, $\nabla f$ is perpendicular to the level curve through the point.
Why it works
Take any curve $\mathbf{r}(t)$ lying in the level surface and passing through $P$. Along it, $F(\mathbf{r}(t)) = k$ is constant, so by the chain rule
$$\frac{d}{dt}F(\mathbf{r}(t)) = \nabla F \cdot \mathbf{r}'(t) = 0.$$
The gradient is perpendicular to the velocity of every curve in the surface through $P$, so it is perpendicular to the whole tangent plane. For a graph $z = f(x, y)$, use $F = f(x, y) - z$: $\nabla F = \langle f_x, f_y, -1 \rangle$, the same normal as in tangent planes.
Worked examples
Worked example: Tangent plane to an ellipsoid
Find the tangent plane to $x^2 + 2y^2 + 3z^2 = 6$ at $(1, 1, 1)$.
- $F = x^2 + 2y^2 + 3z^2$: $\nabla F = \langle 2x, 4y, 6z \rangle = \langle 2, 4, 6 \rangle$.
- ${}2(x - 1) + 4(y - 1) + 6(z - 1) = 0$.
- Divide by 2 and expand: $x + 2y + 3z = 6$.
Answer: $x + 2y + 3z = 6$
Worked example: The normal line
Find the normal line to the same ellipsoid at $(1, 1, 1)$.
- Direction $\nabla F = \langle 2, 4, 6 \rangle$, or the simpler $\langle 1, 2, 3 \rangle$.
- $\mathbf{r}(t) = \langle 1 + t, 1 + 2t, 1 + 3t \rangle$.
Answer: $\mathbf{r}(t) = \langle 1 + t, 1 + 2t, 1 + 3t \rangle$
Worked example: A tangent line to a level curve
Find the tangent line to $x^2 - xy + y^2 = 7$ at $(3, 2)$.
- $\nabla F = \langle 2x - y, -x + 2y \rangle = \langle 4, 1 \rangle$ at $(3, 2)$, a normal to the curve.
- ${}4(x - 3) + 1(y - 2) = 0$, so ${}4x + y = 14$.
Answer: ${}4x + y = 14$
Worked example: A graph as a level surface
Find the tangent plane to $z = x^2 + y^2$ at $(1, 2, 5)$.
- $F = x^2 + y^2 - z$ has $\nabla F = \langle 2x, 2y, -1 \rangle = \langle 2, 4, -1 \rangle$.
- ${}2(x - 1) + 4(y - 2) - (z - 5) = 0$, so $z = 2x + 4y - 5$.
Answer: $z = 2x + 4y - 5$
Common mistake
Using a two-component gradient as a normal in space. For the surface $z = f(x, y)$, the vector $\nabla f = \langle f_x, f_y \rangle$ is perpendicular to level curves in the plane, not to the surface. The surface's normal needs three components: write it as $F = f - z$ and use $\nabla F = \langle f_x, f_y, -1 \rangle$. Check: the point must satisfy your tangent plane equation.