Separability of $\mathbb{C}[x,y_1,\ldots,y_r]$ over $\mathbb{C} + (h,y_1,\ldots,y_r)$

The answer to this MO question says the following:

Lemma 1. Let $$h \in \mathbf C[x]$$ be a polynomial of degree $$n \geq 2$$. Then $$\mathbf C+(h) \subseteq \mathbf C[x]$$ is unramified if and only if $$h$$ is squarefree.

I wonder what happens in higher dimensions, namely:

Question: Is the following claim true: Let $$h \in \mathbb{C}[x]$$ be a polynomial of degree $$n \geq 2$$. Then $$\mathbb{C}[x,y_1,\ldots,y_r]$$ is unramified (equivalently: separable) over $$\mathbb{C}+(h,y_1,\ldots,y_r)$$ if and only if $$h$$ is squarefree.

What I think is that if Lemma 1 can be generalized to the following Lemma 2, then the answer to my current question is positive.

Lemma 2. Let $$A$$ be a commutative $$\mathbb{C}$$-algebra of characteristic zero. Let $$h \in A[x]$$ be a polynomial of degree $$n \geq 2$$. Then $$A +(h) \subseteq A[x]$$ is unramified if and only if $$h$$ is squarefree.

Indeed: If Lemma 2 is true, then $$\mathbb{C}[y_1,\ldots,y_r]+(h) \subset \mathbb{C}[y_1,\ldots,y_r][x]= \mathbb{C}[x,y_1,\ldots,y_r]$$ is unramified if and only if $$h$$ is aquarefree.

Recall a known result about separability (can be found here) which says that given $$R_1 \subseteq R_2 \subseteq R_3$$, if $$R_1 \subseteq R_3$$ is separable, then $$R_2 \subseteq R_3$$ is separable.

Therefore here, if $$h$$ is squarefree, then $$\mathbb{C}[y_1,\ldots,y_r]+(h) \subset \mathbb{C}[x,y_1,\ldots,y_r]$$ is unramified, and then $$\mathbb{C}+ (h,y_1,\ldots,y_r) \subset \mathbb{C}[x,y_1,\ldots,y_r]$$ is unramified.

Any hints and comments are welcome! Thank you.

In Lemma 2, I'm not sure what "squarefree" means. However, the meaning is clear if $$h\in\mathbb{C}[x]$$, and then Lemma 2 is true. Indeed, the inclusion $$\mathbb{C}+(h)\subset \mathbb{C}[x]$$ induces an injection $$A\otimes_\mathbb{C}(\mathbb{C}+(h))\subset A[x]$$. The image is easily checked to be $$A+(h)$$. The lemma follows because being unramified is preserved by base change.