Solve z=x^3y+3x^2-zyx

$$z= { x }^{ 3 } y+3 { x }^{ 2 } -zyx$$

Solve for y

$$\left\{\begin{matrix}y=-\frac{3x^{2}-z}{x\left(x^{2}-z\right)}\text{, }&x\neq 0\text{ and }z\neq x^{2}\\y\in \mathrm{R}\text{, }&z=0\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$x^{3}y+3x^{2}-zyx=z$$

Subtract 3x^{2} from both sides.

$$x^{3}y-zyx=z-3x^{2}$$

Combine all terms containing y.

$$\left(x^{3}-zx\right)y=z-3x^{2}$$

The equation is in standard form.

$$\left(x^{3}-xz\right)y=z-3x^{2}$$

Divide both sides by x^{3}-zx.

$$\frac{\left(x^{3}-xz\right)y}{x^{3}-xz}=\frac{z-3x^{2}}{x^{3}-xz}$$

Dividing by x^{3}-zx undoes the multiplication by x^{3}-zx.

$$y=\frac{z-3x^{2}}{x^{3}-xz}$$

Divide z-3x^{2} by x^{3}-zx.

$$y=\frac{z-3x^{2}}{x\left(x^{2}-z\right)}$$