The answer to this question is negative. A suitable counterexample can be constructed as follows.

Let $Y=\mathbb R$ be the real line with the standard Euclidean topology.
Let $\mathbb Q$ be the subspace of rational numbers in $\mathbb R$. Write $\mathbb R\setminus \mathbb Q$ as the union $\bigcup_{q\in \mathbb Q}X_q$ of pairwise disjoint dense sets in $\mathbb R$. Let $$X=\mathbb Q\oplus\bigoplus_{q\in \mathbb Q}(\{q\}\cup X_q)$$be the topological sum of the spaces $\mathbb Q$ and $\{q\}\cup X_q$ for $q\in\mathbb Q$.

Let $\varphi:X\to Y$ be the natural projection. It is clear that the spaces $X,Y$ are separable, metrizable (and hence Tychonoff) and the function $\varphi:X\to Y$ is skeletal.

On the other hand, in the $\mathbb R$-quotient topology $\tau$ on $Y=\mathbb R$, the set $\mathbb Q$ is closed and nowhere dense in $(Y,\tau)$, witnessing that the function $\varphi:X\to(Y,\tau)$ is not skeletal.

To show that $\mathbb Q$ is closed in $(Y,\tau)$, choose any point $y\in\mathbb R\setminus \mathbb Q$. Find $q\in\mathbb Q$ such that $y\in X_q$. Consider the function $f_q:Y\to\mathbb R$ defined by
$f_q(x)=|x-q|$ if $x\in X_q$ and $f_q(x)=0$, otherwise.

Observe that the composition $f_q\circ \varphi:X\to\mathbb R$ is continuous, which implies that the set $X_q=\{x\in Y:f_q(x)>0\}$ is $\tau$-open, contain $y$, and does not intersect $\mathbb Q$.
This completes the proof of the closedness of $\mathbb Q$.

Assuming that the closed set $\mathbb Q$ is not nowhere dense in $(Y,\tau)$, we can find a nonempty open set $U\subseteq\mathbb Q$. Choose any point $q\in U$ and observe that the preimage $\varphi^{-1}(U)$ is an open set in $X$ containing the point $q\in \{q\}\cup X_q$. Since $\{q\}$ is nowhere dense in $X_q$, the set $\varphi^{-1}(U)$ has nonempty intersection with the set $X_q$ and then $\emptyset \ne \varphi[X_q\cap \varphi^{-1}(U)]\subseteq U\cap X_q\subseteq U\setminus\mathbb Q$, which contradicts the choice of $U\subseteq\mathbb Q$.