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The function $h(x,y) = \frac{20}{3+x^2+2y^2}$ represents the following graph:
Is there anyway to see from the equation, without plotting it, that its graph will look as shown above? Moreover, is there a simple way of coming up with another equation that represents two "mountains", of given heights above the $xy$-plane, next to each other (or even more complicated "mountain ranges" with given properties)?
The function $x^2+y^2+1$ is known to describe a vertical paraboloid, i.e. the result of rotating the parabola $z=x^2+1$ around it axis. Hence it has a single minimum at $(0,0,1)$ and goes to infinity in every direction.
You can dilate the coordinates along the axis $x$ and $y$ to break the rotational symmetry, and every section becomes an ellipse instead of a circle: $z=ax^2+by^2+1$. You can also add a mixed term so that the ellipse axis takes any direction: $ax^2+bxy+cy^2+1$, making sure that $b^2<4ac$ to keep ellipses.
Now the inverse of this function,
$$\frac{z_m}{ax^2+bxy+cy^2+1}$$ is a "mountain-like" function, which has a single maximum at $(0,0,z_m)$ and goes down to zero in every direction. You can adjust the height via the parameter $z_m$, the steepness via $a$ and $c$, and the orientation via $b$.
You can translate a mountain to another location by shifting the coordinates, giving
$$z=\frac{z_m}{a(x-x_m)^2+b(x-x_m)(y-y_m)+c(y-y_m)^2+1}.$$
Now to obtain a more complex landscape, you can combine several mountains by
taking the sum of such functions with different parameters,
taking the maximum.
The sum will result in a blending effect. The maximum keeps the original surfaces and shows them "intersecting". More generally, you can tune the blending by a formula such as
$$\sqrt[\alpha]{z_0^\alpha+z_1^\alpha}$$ where $\alpha$ is a free parameter.
For $r>0$, the equation $x^2+2y^2=r$ is an ellipse centered at the origin. By increasing $r$, the ellipse becomes larger and larger. Moreover the function $$r\to \frac{20}{3+r}$$ is decreasing with respect to $r$. So the given function $f$ attains its maximum value $\frac{20}{3}$ at the origin where $r=0$ (peak of the mountain). Then, as $r$ goes to infinity, the height of the level set (or isoline) of the function, the ellipse $x^2+2y^2=r$, decreases to zero.
In order to have two peaks at height $h>0$, we may try $$f(x,y)=\frac{h}{1+(x^2+y^2)((x-1)^2+(y-1)^2)}.$$ Now the maximum value $h$ is attained at $(0,0)$ and $(1,1)$.
Here are some considerations that can lead you to conclude several aspects of this mountain shape:
If you want multiple mountains, just add some such terms. You probably want to move them by using $x-x_0$ in stread of $x$, and $y-y_0$ instead of $y$. Perhaps you also want to rotate them. So you would go for one of these forms:
$$\frac{a}{b+c(x-x_0)^2+d(y-y_0)^2+e(x-x_0)(y-y_0)}\qquad \frac{a}{b+cx+dy+ex^2+fxy+gy^2}$$
Each of these has $7$ free parameters you can use as knobs to tweak. The left version makes it easier to tell that the center is at $(x_0,y_0)$ while the right version with $\dots,e,f,g$ has a simpler structure in the denominator. In both versions, you can scale all numbers by a constant factor to achieve the same shape, so geometrically there are only $6$ degrees of freedom in picking individual mountains according to this pattern.
If you "stretch" the graph in the $y$ direction, that is, replace $y$ with $\frac{1}{\sqrt{2}}y$, then the function becomes $$g(x,y) = \frac{1}{3+x^2+y^2}.$$
So basically, the graph of the original function is the graph of $g$, but stretched along the $y$ axis by a factor of $\frac{1}{\sqrt{2}}$.
The graph of $g$ is simpler, since $g(x,y)=\frac{1}{3+\|(x,y)\|^2}$. This (3D) graph is just the (2D) graph of $\frac{1}{3+x^2}$, rotated around the $y$ axis.
$$h(x,y) = \frac{20}{3+x^2+2y^2}$$ Obviously $h$ is maximum when $x=0$ and $y=0$. So it appears as a peak.
$h\to 0$ when $x\to\infty$ and/or $y\to\infty$. So it appears as plains.
More generally : $$h(x,y) = \frac{H}{1+A(x-a)^2+B(y-b)^2}$$ $H>0,A>0,B>0,a,b$ are constants.
The coordinates of the peak are $(a,b)$ and its high is $H$.
$A$ and $B$ define the size of the ellipse as shape of the "mountain".
Several "mountains" are obtained for example with : $$h(x,y) = \frac{H_{1}}{1+A_1(x-a_1)^2+B_1(y-b_1)^2}+ \frac{H_{2}}{1+A_2(x-a_2)^2+B_2(y-b_2)^2}+\text{etc.}$$
With some imagination, yes, there is a quick and dirty way.
Know how the function $y=\frac{1}{x^2}$ looks like and know how the function of an ellipse in the $x,y-$plane and then try to analyze the intersections of the $3D$ surface $z=\frac{20}{3+x^2+2y^2}$ with the $y,z-$ and $x,z-$planes.
(1) Assume $x=0$, then we are in the $y,z-$plane and we have $z = \frac{20}{3+2y^2}$.
We can just draw $\color{red}{z=\frac{1}{y^2}}$ instead of $z = \frac{20}{3+2y^2}$ to get a rough idea about the intersection of the surface with the $y,z-$plane:
(2) Assume $y=0$, then we are in the $x,z-$plane and we have $z = \frac{20}{3+x^2}$.
We can just draw $\color{blue}{z=\frac{1}{x^2}}$ instead of $z = \frac{20}{3+x^2}$ to get a rough idea about the intersection of the surface with the $x,z-$plane:
(3) Now at the point $(x,y)=(0,0)$ we have $z(0,0) = \frac{20}{3}$. So, the point $(x,y,z)=(0,0,20/3)$ is on the surface:
(4) We see that the surface doesn't touch the $x,y-$plane. That's why we could try $z=1$ instead of $z=0$ to see the result of the intersection with the plane $z=1$ (which has points parallel to the $x,y-$plane). Setting $z=1$ and plugging in the function yields $\color{green}{x^2+2y^2=17}$, which is an equation of an ellipse and we are done
Some hints regarding graphs of $ z= f(x,y) $ Monge forms: In $$ z=\frac {1}{r^2}= \frac {1}{x^2+y^2} $$ it goes infinite at pole $r=0$ of a single peak.
To remedy this i.e., to have a flat plateau add some positive real number in denominator.. like
$$ z=\frac {1}{4+r^2}= \frac {1}{4+ x^2+y^2} $$
or Gauss probability /Bell curve rotated peak where the finite polynomial series has even terms included in exponential function to force center to become flat.
$$ z= e^ {- (x^2+y^2) / (2 \sigma^2)} $$
To get two peaks work around two foci. For example Cassinian Ovals of const. product $b^4$, height $z$ can replace its single parameter/constant $b^4$.
$$ b^4 = (x+a)^2 * (x-a)^2 \rightarrow z^4 = (x+a)^2 * (x-a)^2 $$
We can raise the power of polynomial to get infinitely many peaks, valleys and cols. Sine function for example produces infinite periodic mountain range.. $$ z = \sin x + \sin y $$
