Solve z=ysqrt{x}-y^2-x+6y

$$z = y \sqrt{ x } - { y }^{ 2 } -x+6y$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=0\text{, }&z=y\left(6-y\right)\\x=\frac{\left(\sqrt{-3y^{2}+24y-4z}+y\right)^{2}}{4}\text{, }&arg(\frac{\sqrt{-3y^{2}+24y-4z}+y}{2})<\pi \\x=\frac{\left(-\sqrt{-3y^{2}+24y-4z}+y\right)^{2}}{4}\text{, }&arg(\frac{-\sqrt{-3y^{2}+24y-4z}+y}{2})<\pi \end{matrix}\right.$$

Solve for y (complex solution)

$$y=\frac{-\sqrt{-3x-4z+12\sqrt{x}+36}+\sqrt{x}+6}{2}$$

$$y=\frac{\sqrt{-3x-4z+12\sqrt{x}+36}+\sqrt{x}+6}{2}$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\left(\sqrt{-3y^{2}+24y-4z}+y\right)^{2}}{4}\text{, }&\frac{\sqrt{-3y^{2}+24y-4z}+y}{2}\geq 0\text{ and }z\leq -\frac{3y^{2}}{4}+6y\\x=\frac{\left(-\sqrt{-3y^{2}+24y-4z}+y\right)^{2}}{4}\text{, }&\frac{-\sqrt{-3y^{2}+24y-4z}+y}{2}\geq 0\text{ and }z\leq -\frac{3y^{2}}{4}+6y\end{matrix}\right.$$

Solve for y

$$y=\frac{-\sqrt{-3x-4z+12\sqrt{x}+36}+\sqrt{x}+6}{2}$$

$$y=\frac{\sqrt{-3x-4z+12\sqrt{x}+36}+\sqrt{x}+6}{2}\text{, }x\geq 0\text{ and }z\leq -\frac{3x}{4}+3\sqrt{x}+9$$