Is there a way to draw 3d graphs on paper? I want to draw three dimensional graphs on paper. How would I do this though? Would I have to fold paper? Or would I have to rotate each point projected onto a 2d plane to make it suitable for my view?
By 3d graphs, I mean $z=f(x,y)$
I can't stress enough that I HAVE to use paper and pencil
 A: Suppose you want to plot $z = f(x, y)$ over the rectangle $[a, b] \times [c, d]$, i.e., for $a \leq x \leq b$ and $c \leq y \leq d$, using a mesh grid of size $m \times n$. One simple approach is to use "orthogonal projection":


*

*Select a function and the rectangle over which you want to plot. Find or estimate the minimum and maximum values the function achieves.

*Tack down a sheet of paper on a drafting board. Using a T-square, $30$-$60$-$90$ triangle, and ruler, lay out a parallelogram on your paper with one side horizontal, the rectangular domain seen in perspective, and mark off the subdivision points along the outer edges ($m$ equal intervals in the $x$-direction and $n$ equal intervals in the $y$-direction). Use the minimum and maximum values of the function to estimate where on the paper the domain should be drawn, and to decide on the overall vertical scale of the plot.
For definiteness (see diagram below), let's call the bottom edge of the parallelgram $x = x_{0}$ and the left edge $y = y_{0}$. (Depending on how the parallelogram is oriented, you might have $x_{0} = a$ or $b$, and $y_{0} = c$ or $d$.) Using a sharp 6H pencil, subdivide the parallelogram (the domain) into an $m \times n$ grid.

*Calculate the step sizes
$$
\Delta x = \frac{b - a}{m},\qquad
\Delta y = \frac{d - c}{n}.
$$
(The formulas below assume the step sizes are positive, i.e., thet $x_{0} = a$ and $y_{0} = c$. The modifications should be fairly obvious if $x$ decreases from bottom to top and/or $y$ decreases from left to right.)

*To effect hidden line removal, we'll plot front to back. Calculate the "front row" values
$$
f(x_{0}, y_{0} + j\, \Delta y),\qquad 1 \leq j \leq n.
$$
Locate each point $(x_{0}, y_{0} + j\, \Delta y)$ in your grid, measure up or down to the appropriate height, and put a dot at that location. When you're plotted these $n$ points, connect the dots from left to right with a 2B pencil.

*Now iterate the following step, letting $i$ run from $1$ to $m$. Calculate the values
$$
f(x_{0} + i\, \Delta x, y_{0} + j\, \Delta y),\qquad 1 \leq j \leq n.
$$
Locate each point $(x_{0} + i\, \Delta x, y_{0} + j\, \Delta y)$ in your grid, and measure up or down to the appropriate height. When you're plotted these $n$ points:


*

*Draw one row of "front-to-back" segments: For each $j = 1, \dots, n$, connect the dot over $(x_{0} + (i - 1)\, \Delta x, y_{0} + j\, \Delta y)$ to the dot over $(x_{0} + i\, \Delta x, y_{0} + j\, \Delta y)$. (Use light lines or no lines if the segment lies behind a part of the surface you have already plotted.)

*Draw the $i$th row: For each $j = 1, \dots, n$, connect the dot over $(x_{0} + i\, \Delta x, y_{0} + j\, \Delta y)$ to the dot over $(x_{0} + i\, \Delta x, y_{0} + (j + 1)\, \Delta y)$. (Again, use light lines or no lines if the segment lies behind a part of the surface you have already plotted.)
Speaking from experience, the process takes (with a calculator) about eight hours for a $20 \times 20$ mesh. It's doubtless faster to tabulate all the values of $f$ at the mesh points, then to plot points by reading from the table. (I was impetuous as a student, and alternately calculated one value and plotted one point.)
The diagram shows (a computer-drawn version of) the first function I plotted, shifted up to avoid overlap with the rectangular mesh in the domain. When you're actually plotting on paper, you probably don't want to waste the vertical space, and so will have to draw the grid lightly and plot over it.

