$0^\#\text{ exists}$
This statement is upwards absolute. If $0^\#$ exists then it exists in all larger [transitive] models. However it is not downwards absolute because $L\models\lnot\exists 0^\#$.
Equally if the universe is "sufficiently different" from some of its inner models then the formula $\exists f\colon\alpha\to\beta\text{ a bijection}$ can be true in some models and false in others.
For example if $\omega_1^L$ is countable in $M$ then $M\models\exists f\colon\omega\to\omega_1^L$, and any model extending $M$ also satisfies this formula, however $L$ itself doesn't. Note that $\omega$ is definable without parameters and $\omega_1^L$ is the least ordinal that has no constructible injection into $\omega$, and so it is also definable without parameters.
An even simpler variant of the previous example is simply writing the axiom $V\neq L$. Recall that there is a formula $\varphi(x)$ such that $\varphi(x)$ is true if and only if $x\in L$. Therefore $V=L$ can be expressed as $\forall x.\varphi(x)$.
Writing $\exists x.\lnot\varphi(x)$ is clearly not downwards absolute, because this statement would be false in $L$ of the model; but it is also clearly upwards absolute because if $M\subseteq N$ (both transitive with the same ordinals) and $M\models V\neq L$ then $N$ cannot have satisfy $V=L$.