one at the origin and the second at point A. Determine the Coordinates of A
I was going to try to solve it but after searching it up it looks like I have to use partial differentiation which I haven't studied. Is this true? Also, I still want to learn how to do it either way so how would you do it?
The only reasonable approach to this problem that I know of is to use partial differentiation. A nice explanation of partial derivatives is given here.
The stationary points of a function are the points where all of its partial derivatives are equal to zero. In this case, we compute the partial derivatives of $z$ to be $$ \frac{\partial z}{ \partial x} = 3x^2 - 6y, \quad \frac{\partial z}{\partial y} = 3y^2 - 6x. $$ Setting these equal to zero gives us the system of equations $$ 3x^2 = 6y, \quad 3y^2 = 6x. $$ Solving the first equation yields $y = \frac 12 x^2$. Substituting this into the second equation yields $$ 3[\frac 12 x^2]^2 = 6x \implies \frac 34 x^4 = 6x \implies x^4 = 8x \implies x^4 - 8x = 0 \implies\\ x(x^3 - 8) = 0. $$ One solution to this equation is $x = 0$, which yields the corresponding $y$-coordinate $y = \frac 12 0^2 = 0$. The other solution is $x = \sqrt[3]{8} = 2$, which yields the corresponding $y$-coordinate $\frac 12 2^2 = 2$.
So, the coordinates of $A$ are $(2,2)$.