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In Ian N. Sneddon's Elements of Partial Differential Equations, there is the following text:

If the rectangular Cartesian co-ordinates of any point $P(x,y,z)$ in space satisfies the relation $$f(x,y,z)=0\tag1$$ then the point $P$ is said to lie on a surface, whose equation is given by $(1)$.

This implies that the equation of a surface is of the form $f(x,y,z)=0$.

But it does not mention why it is so. It simply states that this is obvious.

However, I want to know if there is any proof of the fact that equations of this form are surfaces, or whether it is just a definition?

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    Is continuity of $f$ assumed? Does it come down to whether $\{(x,y,z) \mid f(x,y,z)=0\}$ contains an open ball? – 2017-02-05
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    No idea. It doesn't say anything about this. – 2017-02-05

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I think the statement is meant to be taken intuitively and not formally in the sense that if you impose a "reasonable" equation on the coordinates $(x,y,z)$ you "expect" to get an object that has one dimension less and so deserves to be called a "surface".

In general, you wouldn't want to define a surface as a solution set $$ S = \{ (x,y,z) \in \mathbb{R}^3 | \, f(x,y,z) = 0 \} $$

without imposing some conditions on $f$. If we allow arbitrary $f$ then we might as well allow $f \equiv 0$ but then $S = \mathbb{R}^3$ is three-dimensional so it doesn't really deserve the name surface. Or we can take $f = x^2 + y^2 + z^2$ and then $S = \{ (0,0,0) \}$ is a point which is again, not very satisfactory.

One way to make this rigorous is to define a surface as a two-dimensional embedded smooth submanifold of $\mathbb{R}^3$. Then the inverse function theorem will tell you that if $f$ is smooth and $0$ is a regular value of $f$ then $S$ is a surface.