Step 1: Recall what odd means.
A function is odd when flipping the input sign flips the whole output sign, that is $f(-x)=-f(x)$. A function is even when $f(-x)=f(x)$. We will test which rule our function obeys.
Step 2: Write the function.
We are given $f(x)=\sin\left(\log\left(x+\sqrt{x^2+1}\right)\right)$. The tricky part is the inside expression $x+\sqrt{x^2+1}$.
Step 3: Replace $x$ with $-x$.
Then $f(-x)=\sin\left(\log\left(-x+\sqrt{x^2+1}\right)\right)$. Notice the square root part stays the same because $(-x)^2=x^2$.
Step 4: Use a neat identity.
Multiply $(x+\sqrt{x^2+1})$ by $(\sqrt{x^2+1}-x)$. This gives $(x^2+1)-x^2=1$. So the two expressions are reciprocals: $\sqrt{x^2+1}-x=\frac{1}{x+\sqrt{x^2+1}}$.
Step 5: Turn the reciprocal into a minus sign inside the log.
Since $\log\left(\frac{1}{A}\right)=-\log A$, we get $\log\left(-x+\sqrt{x^2+1}\right)=-\log\left(x+\sqrt{x^2+1}\right)$.
Step 6: Use the odd nature of sine.
So $f(-x)=\sin\left(-\log\left(x+\sqrt{x^2+1}\right)\right)$. Because $\sin(-\theta)=-\sin\theta$, this equals $-\sin\left(\log\left(x+\sqrt{x^2+1}\right)\right)=-f(x)$.
Step 7: Conclude.
Since $f(-x)=-f(x)$, the function is odd.
\[ \boxed{\text{An odd function}} \]