The answer ultimately depends on what level of rigor the situation calls for. In a typical multivariable calculus course, for instance, you may usually take for granted that polynomials (in any number of variables) and $f(z) = \ln(z)$ are both continuous functions on their domains, whereas this may or may not be the case in a course on foundational real analysis. So if you're in the former camp (as your tags suggest) then you can probably get away with just stating that the relevant polynomial in $x$ and $y$, $f(z) = \ln z$, and compositions of continuous functions are all continuous.
Now for the fun part: How do you actually show these things with some rigor? Again, that will depend on where you start/how much you can assume, but here are some off-the-cuff answers for how to do this with a reasonable toolset at your disposal. I've left the $(\epsilon,\delta)$ arguments in a form that can probably be made cleaner but that gives you an idea of how to cook these sorts of proofs up yourself.
Claim: Every polynomial $P(x,y)$ is continuous on $\mathbb{R}^2$.
Proof: Note first that $q_C(x,y) = C$ for any constant $C$, $P_1(x,y) = x$, and $P_2(x,y) = y$ are all continuous functions. (Using the usual $(\epsilon,\delta)$ definition of continuity at a point, given $\epsilon > 0$ set $\delta = \epsilon$.) Since products and sums of continuous functions are continuous (by the same argument used in one variable) and $P(x,y)$ is a sum of products of functions of the form $q_C$ (for varying $C$), $P_1$, and $P_2$), it follows that $P$ is continuous.
Claim: $f(z) = \ln(z)$ is continuous at every $z > 0$
Proof: The usual formal definition of $\ln(z)$ for $z > 0$ is
$$
\ln(z) = \int_1^z \frac{1}{t}\,dt.
$$
Since $1/t$ is defined and Riemann integrable on $[1,z]$ (or $[z,1]$ for $0<z<1$) one can show that $\ln(z)$ is continuous.
Claim: Compositions of functions are continuous
Suppose $A \subset \mathbb{R}^m$, $B \subset \mathbb{R}^n$, and $C \subset \mathbb{R}^p$.
Suppose $f: A \to \mathbb{R}^n$ is continuous at $a$ with $f(a) \in B$ and $g: B \to C$ is continuous at $f(a)$. Then we must show that $h = g \circ f$ is continuous at $a$.
Let $\epsilon > 0$. Since $g$ is continuous at $f(a)$, there exists $\delta_1 = \delta_1(\epsilon) > 0$ such that if $y \in B$ and $\|y-f(a)\| < \delta_1$ then $\|g(y) - f(a)\| < \epsilon$.
When considering $f$, think of $\delta_1$ as taking the role of your given $\epsilon$.
Since $f$ is continuous at $a$, there exists $\delta > 0$ such that if $x \in A$ and $\|x-a\| < \delta$ then $\|f(x) - f(a)\| < \delta_1$.
Thus for all $x \in A$ with $\|x-a\| < \delta$ we have that $f(x)$ is a point in $B$ at which $\|f(x)-f(a)\| < \delta_1$, and so by the above statement about the continuity of $g$ at $f(a)$ we conclude $\|h(x) - h(a)\| = \|g(f(x)) - g(f(a))\| < \epsilon$.
