Let $z = g(x,y) = xy-8x-y^2+12y+160$. I would like to graph some level curves of the surface.
So set $z = a$ be a constant. I then have $$a = xy - 8x - (y^2-12y-160) = xy - 8x - (y^2-12y+36 - 196) = xy-8x - (y-6)^2+196.$$ So the equation is
$$(y-6)^2-x(y-6)+2x=(y-6)^2 + x(8-y)=(y-6)^2 +8x - xy = 196 - a$$ which I still cannot see what type of $xy-$curve it is.
But $$(y-6)^2 - x(y-6) + \frac{x^2}{4} - \frac{x^2}{4} + 2x = (y-6-\frac{x}{2})^2 - (\frac{x^2}{4}+ (2) (\frac{x}{2})(2)+4 -4) = (y-6-\frac{x}{2})^2 - (\frac{x}{2}+2)^2+4.$$
Hence, the curve form is $$(y-6-\frac{x}{2})^2 - (\frac{x}{2}+2)^2 = 192-a$$ which I still not sure how this curve looks like (without using computer or graphing calculator).
I just wonder if there is a way to algebraically manipulate the expression to match usual equation of curve in $xy-$plane, or the only chance is to use graphic software to plot it.
If you just want to picture it, there are some shape preserving techniques you can use.
Suppose you have $z=f(x,y)=Ax^2+By^2+Cx+Dy+Exy+F=z_0$
First do a change of variables to get rid of the $xy$ term.
$$x=x'\cos\theta+y'\sin\theta$$ $$y=-x'\sin\theta+y'\cos\theta$$
This has the effect of rotating the figure about the z-axis by an angle $\theta$. Swap in these values for $x$ and $y$, and you get a new expression in terms of $x'$ and $y'$ with new coefficients. Now you have a $x'y'$ cross term that you can set equal to zero by selecting the necessary value for $\theta$.
Afterwards, your equation becomes:
$$A'x'^2+B'y'^2+C'x'+D'y'+F'=z_0$$
Now you want to consolidate terms of x' together, and terms of y' together:
$$ A'(x'^2+\frac{C'}{A'}x)+B'(y'^2+\frac{B'}{D'})+F'=z_0 $$
Now complete the square and subtract $F'$ from both sides:
$$A'(x'+\frac{C'}{2A'})^2+B'(y'+\frac{B'}{2D'})^2=z_0+\frac{C'^2}{4A'^2}+\frac{B'^2}{4D'^2}-F'=z_0'$$
Now divide by $z0'$:
$$\frac{(x'+\frac{C'}{2A'})^2}{z_0'/A'}+\frac{(y'+\frac{B'}{2D'})^2}{z_0'/B'}=1$$
If the denominators are both positive, you have an ellipse, a circle if they are equal. IF they have opposite signs, you have a hyperbola. If only one of $A'$ or $B'$ are zero, then you have a parabola. If both are zero, your level curve is a line.
The constant terms within the parentheses tell you where the center of the figure is. The denominators give you some information about scaling relative the "square" conic sections. The angle figured out previously tells you how to rotate these figures back to the original.