Evaluate \(\displaystyle z_x\) at \(\displaystyle (1, 2, 3)\) if \(\displaystyle x^3+y^3+z^3+6xyz=72\)

Part of me says I must solve this in terms of z in order to find \(\displaystyle z_x\) but I can't see how.

The other part thinks I can just find \(\displaystyle z_x\) implicitly and plug in the point but I am getting an answer that doesn't make any sense.

What should I do?

We assume that \(\displaystyle z\) is a function of \(\displaystyle x\), so you need to use the Chain Rule...

\(\displaystyle x^3 + y^3 + z^3 + 6xyz = 72\)

\(\displaystyle \frac{\partial}{\partial x}(x^3 + y^3 + z^3 + 6xyz) = \frac{\partial}{\partial x}(72)\)

\(\displaystyle \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial x}(y^3) + \frac{\partial}{\partial x}(z^3) + \frac{\partial}{\partial x}(6xyz) = 0\)

\(\displaystyle 3x^2 + 0 + \frac{\partial z}{\partial x}\,\frac{\partial}{\partial z}(z^3) + 6y\,\frac{\partial}{\partial x}(xz) = 0\) (Chain Rule)

\(\displaystyle 3x^2 + 3z^2\,\frac{\partial z}{\partial x} + 6y\left[x\,\frac{\partial}{\partial x}(z) + z\,\frac{\partial}{\partial x}(x)\right] = 0\) (Product Rule)

\(\displaystyle 3x^2 + 3z^2\,\frac{\partial z}{\partial x} + 6y\left[x\,\frac{\partial z}{\partial x} + z(1)\right] = 0\)

\(\displaystyle 3x^2 + 3z^2\,\frac{\partial z}{\partial x} + 6xy\,\frac{\partial z}{\partial x} + 6yz = 0\)

\(\displaystyle 3z^2\,\frac{\partial z}{\partial x} + 6xy\,\frac{\partial z}{\partial x} = -(3x^2 + 6yz)\)

\(\displaystyle (3z^2 + 6xy)\frac{\partial z}{\partial x} = -(3x^2 + 6yz)\)

\(\displaystyle \frac{\partial z}{\partial x} = -\frac{3x^2 + 6yz}{3z^2 + 6xy}\)

\(\displaystyle \frac{\partial z}{\partial x} = -\frac{x^2 + 2yz}{z^2 + 2xy}\).

Now substitute \(\displaystyle (x, y, z) = (1, 2, 3)\).