How does this surface looks like? The formula is $({x^4 \over{a^4}} + {y^2 \over{b^2}} + {z^2 \over{c^2}})^2 = {x^2 \over{p^2}}$, where $a,b,c,p$ is positive. The left part is pretty similuar to ellipsoid, but there are few differences. I also have tried to use the difference of two squares and brackets expansion, but it didn't help. 
And the second function - what is the volume of this body?
Thanks for any help!

As a first step, move the right-hand side to the left side, and use the conjugate rule. You will get
$$
\Bigl(\frac{x^4}{a^4}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{x}{p}\Bigr)\cdot
\Bigl(\frac{x^4}{a^4}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-\frac{x}{p}\Bigr)=0
$$
You will in general get two "blobs", symmetric with respect to $x$. In the case $a=b=c=p=1$, it looks like this:

To calculate the volume, we can consider $x\geq 0$ (and multiply the result by $2$). The only parenthesis that can be zero is the second,
$$
\Bigl(\frac{x^4}{a^4}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-\frac{x}{p}\Bigr)
$$
and if we cut it with planes orthogonal to the $x$-axis, we get ellipses,
$$
\frac{y^2}{b^2}+\frac{z^2}{c^2}=\frac{x}{p}-\frac{x^4}{a^4}
$$
for $0<x<a^{4/3}/p^{1/3}$ and nothing for other planes. The ellipses have half-axes
$$
b\sqrt{x/p-x^4/a^4}\quad \text{and}\quad c\sqrt{x/p-x^4/a^4}
$$
so the area $A(x)$ of each such ellips equals
$$
A(x)=\pi b\sqrt{x/p-x^4/a^4}c\sqrt{x/p-x^4/a^4}=\pi bc(x/p-x^4/a^4).
$$
Thus, the volume $V$ of both blobs equals (I leave it to you to perform the integral)
$$
V=2\int_0^{a^{4/3}/p^{1/3}}A(x)\,dx=\cdots=\frac{3\pi a^{8/3}bc}{5p^{5/3}}.
$$
