$\frac{dz}{dx}$ term explanation for $x^3 + y^3 + z^3 +6xyz = 1$

Question

I know that the full answer is $$3x^2 + 3z^2 \cdot \frac{dz}{dx} +6yz + (6xy \cdot \frac{dz}{dx})$$ but where does the final part in brackets come from?

I know how to solve the rest, but an explanation of the final term would be greatly appreciated!

Answer 1

It is the chain rule, and we get $$3x^2+3z^2z'(x)+6yz+6xyz'(x)=0$$

Answer 2

We have that assuming $y$ constant

$$x^3 + y^3 + z^3 +6xyz = 1 \implies 3x^2dx+3z^2dz+6yzdx\color{red}{+6xydz}=0$$

that is dividing by $dx$

$$3x^2+3z^2\frac{dz}{dx}+6yz\color{red}{+6xy\frac{dz}{dx}}=0\implies \frac{dz}{dx}=-\frac{3x^2+6yz}{3z^2dz+6xy}$$