Consider the first order language $\mathcal{L}=\{\in,\subseteq\}$ and $\{\in\}$-theory $\text{ZFC}$. Is there a formula $\psi (x,y) \in \{\subseteq\}-Form$ with the following property?
$ZFC\cup\{\forall x\forall y~(x\subseteq y\leftrightarrow \forall z (z\in x\rightarrow z\in y))\}\vdash$$ ~\forall x\forall y~(x\in y\leftrightarrow \psi (x,y))$
The edited question is much more interesting than the original.
The answer is no, you cannot define $\in$ from $\subseteq$. To see this, note from this previous MO answer that there are nontrivial automorphisms of $\langle V,\subseteq\rangle$. Namely, let $f:V\to V$ be any permutation, and let $\pi(x)=f[x]$, the pointwise image of $x$ under $f$. It follows that $x\subseteq y\iff \pi(x)\subseteq\pi(y)$. But we may easily choose $f$ so that $\pi$ does not respect $\in$, that is, so that for some $x,y$ we have $x\in y$ but $\pi(x)\notin \pi(y)$. For example, let $f$ swap $\emptyset$ with some other set nonempty set $a$, and let $x=\emptyset$ and $y=\{\emptyset\}$, so that $x\in y$, but $\pi(x)=f[\emptyset]=\emptyset$ and $\pi(y)=f[\{\emptyset\}]=\{f(\emptyset)\}=\{a\}$ so that $\pi(x)\notin\pi(y)$.
It follows that we cannot define $\in$ from $\subseteq$, since $\psi(x,y)\iff \psi(\pi(x),\pi(y))$ for any formula using only $\subseteq$, since it is an $\subseteq$-automorphism.
