In Calculus 1 you found slopes of curves like $x^3 + y^3 = 6xy$ by implicit differentiation. The multivariable chain rule turns that process into a one-line formula, and it extends to surfaces given implicitly.
Key idea
If $F(x, y) = 0$ defines $y$ as a function of $x$ (and $F_y \ne 0$):
$$\frac{dy}{dx} = -\frac{F_x}{F_y}.$$
If $F(x, y, z) = 0$ defines $z$ as a function of $x$ and $y$ (and $F_z \ne 0$): $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$ and $\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}$.
Why it works
Along the curve, $F(x, y(x))$ stays equal to ${}0$, so its derivative with respect to $x$ is ${}0$. By the chain rule with the two paths $F \to x$ and $F \to y \to x$:
$$F_x \cdot 1 + F_y\frac{dy}{dx} = 0.$$
Solve for $\frac{dy}{dx}$: the minus sign comes from moving $F_x$ to the other side. For surfaces, hold $y$ fixed and do the same with $z(x, y)$: $F_x + F_z\frac{\partial z}{\partial x} = 0$.
Worked examples
Worked example: A slope on a curve
Find $\frac{dy}{dx}$ for $x^3 + y^3 = 6xy$ at $(3, 3)$.
- $F = x^3 + y^3 - 6xy$: $F_x = 3x^2 - 6y$, $F_y = 3y^2 - 6x$.
- At $(3, 3)$: $F_x = 27 - 18 = 9$ and $F_y = 27 - 18 = 9$.
- $\frac{dy}{dx} = -\frac{9}{9} = -1$.
Answer: $-1$
Worked example: As an expression
Find $\frac{dy}{dx}$ for $x^2 + xy + y^2 = 7$.
- $F = x^2 + xy + y^2 - 7$: $F_x = 2x + y$ and $F_y = x + 2y$.
- $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$, the same as differentiating both sides with respect to $x$ and solving.
Answer: $-\frac{2x + y}{x + 2y}$
Worked example: A surface
The sphere $x^2 + y^2 + z^2 = 9$ defines $z$ near $(1, 2, 2)$. Find $\frac{\partial z}{\partial x}$ there.
- $F = x^2 + y^2 + z^2 - 9$: $F_x = 2x$ and $F_z = 2z$.
- $\frac{\partial z}{\partial x} = -\frac{2x}{2z} = -\frac{x}{z} = -\frac{1}{2}$ at the point.
- Check: on the top half $z = \sqrt{9 - x^2 - y^2}$, and $z_x = \frac{-x}{\sqrt{9 - x^2 - y^2}} = -\frac{1}{2}$.
Answer: $-\frac{1}{2}$
Common mistake
Forgetting the minus sign. It is easy to write $\frac{F_x}{F_y}$. Test on a circle $x^2 + y^2 = 25$ at $(3, 4)$: the slope must be negative there (the top right of the circle slopes down), and $-\frac{F_x}{F_y} = -\frac{6}{8} = -\frac{3}{4}$ is. Also keep the right variable in the denominator: $F_z$ for $\frac{\partial z}{\partial x}$, $F_y$ for $\frac{dy}{dx}$.