[proofplan] Both claims follow from the multivariate chain rule for polynomial differentiation, applied on the linear subspace $T_{V,p} \subset k^n$. For part (a), we first show $d\varphi_p$ lands in $T_{W, \varphi(p)}$ by observing that for $h \in I(W)$, the function $h \circ \varphi$ vanishes on the open set $U$ and hence (by irreducibility of $V$) lies in $I(V)$, whose gradients kill $T_{V,p}$ by definition. Then well-definedness follows from the same idea: two polynomial expressions of $\varphi$ differ by elements of $I(V)$, whose gradients also kill $T_{V,p}$. For part (b), we expand both sides as matrix products and apply the standard chain rule identity $J(\psi \circ \varphi)(p) = J\psi(\varphi(p)) \cdot J\varphi(p)$, then restrict to $T_{V,p}$. [/proofplan]

[step:Fix polynomial representatives of $\varphi$ near $p$] On the open set $U \ni p$ where $\varphi$ is defined, the polynomial representation $\varphi|_U = (f_1, \ldots, f_N)$ is given by hypothesis. (A rational map $\varphi : V \dashrightarrow W$ is a ratio $(f_1 / g, \ldots, f_N / g)$ of polynomials with common denominator $g$ nonvanishing on $U$; on the open set $\{g \neq 0\}$ we absorb the denominator by working with the representatives $f_j / g$ viewed as elements of the localisation $k[X_1, \ldots, X_n][1/g]$, and since $g(p) \neq 0$ each partial $\partial (f_j / g) / \partial X_i$ evaluates to a well-defined element of $k$ at $p$. For notational cleanliness we write $f_j$ for these representatives; the polynomial chain rule below applies verbatim to rational functions whose denominators are nonvanishing at $p$.) The Jacobian \begin{align*} J\varphi(p) = \left( \frac{\partial f_j}{\partial X_i}(p) \right)_{1 \leq j \leq N,\ 1 \leq i \leq n} \in k^{N \times n} \end{align*} is well-defined as a matrix over $k$, and $d\varphi_p : k^n \to k^N$ is defined as $v \mapsto J\varphi(p) \cdot v$. We will show that the restriction to $T_{V, p}$ lands in $T_{W, \varphi(p)}$ and is independent of the choice of representatives. [/step]

[step:Show $d\varphi_p$ lands in $T_{W, \varphi(p)}$] Take $h \in I(W) \subset k[Y_1, \ldots, Y_N]$. Since $\varphi$ maps $U \subset V$ into $W$, the composition $h \circ \varphi = h(f_1, \ldots, f_N)$ vanishes on $U$. The set $U$ is a nonempty Zariski-open subset of the irreducible variety $V$, hence dense in $V$, so a polynomial (or, more generally, a regular function on $U$) that vanishes on $U$ vanishes on all of $V$ as a regular function. Hence \begin{align*} h \circ \varphi \in I(V). \end{align*} Apply the multivariate chain rule for polynomial differentiation — a purely algebraic identity in $k[X_1, \ldots, X_n]$ that follows from the Leibniz rule and linearity of $\partial / \partial X_i$: \begin{align*} \frac{\partial (h \circ \varphi)}{\partial X_i}(p) = \sum_{j=1}^{N} \frac{\partial h}{\partial Y_j}(\varphi(p)) \cdot \frac{\partial f_j}{\partial X_i}(p), \end{align*} equivalently as row vectors, \begin{align*} \nabla(h \circ \varphi)(p) = \nabla h(\varphi(p)) \cdot J\varphi(p). \end{align*} For $v \in T_{V,p}$, since $h \circ \varphi \in I(V)$, the defining condition $\nabla g(p) \cdot v = 0$ for all $g \in I(V)$ gives \begin{align*} 0 = \nabla(h \circ \varphi)(p) \cdot v = \nabla h(\varphi(p)) \cdot (J\varphi(p) \cdot v). \end{align*} Since this holds for every $h \in I(W)$, the vector $J\varphi(p) \cdot v \in k^N$ is annihilated by every $\nabla h(\varphi(p))$ with $h \in I(W)$, which is exactly the condition $J\varphi(p) \cdot v \in T_{W, \varphi(p)}$. Hence $d\varphi_p$ maps $T_{V, p}$ into $T_{W, \varphi(p)}$. [/step]

[step:Verify independence of the polynomial presentation of $\varphi$]Suppose $\varphi$ is also given by an $N$-tuple $(\tilde f_1, \ldots, \tilde f_N)$ on a (possibly different) open neighbourhood $\tilde U \ni p$ in $V$. On $U \cap \tilde U$ (a nonempty open subset of $V$), $f_j$ and $\tilde f_j$ represent the same $j$-th coordinate of $\varphi$, hence take the same values as regular functions. Therefore \begin{align*} f_j - \tilde f_j \end{align*} is a regular function on $V$ (after clearing common denominators, an element of $k[X_1, \ldots, X_n]/I(V)$) vanishing on $U \cap \tilde U$. By irreducibility of $V$, $U \cap \tilde U$ is dense in $V$, so $f_j - \tilde f_j$ vanishes on all of $V$: \begin{align*} f_j - \tilde f_j \in I(V) \quad \text{for } j = 1, \ldots, N. \end{align*} By the same calculation as in Step 2, for any $g \in I(V)$, $\nabla g(p) \cdot v = 0$ for all $v \in T_{V, p}$. Applying this with $g = f_j - \tilde f_j$: \begin{align*} \nabla(f_j - \tilde f_j)(p) \cdot v = 0 \quad \text{for all } v \in T_{V, p}. \end{align*} Hence the $j$-th rows of $J\varphi(p)$ and $J\tilde\varphi(p)$ agree as linear functionals on $T_{V, p}$. This holds for each $j = 1, \ldots, N$, so the two Jacobians $J\varphi(p)$ and $J\tilde\varphi(p)$ agree as linear maps $T_{V, p} \to k^N$, i.e. $d\varphi_p = d\tilde\varphi_p : T_{V, p} \to T_{W, \varphi(p)}$.[/step]

[guided]The concern about well-definedness is real: $\varphi$ is given by polynomial coordinates only modulo the relations $I(V)$, and different polynomial choices for the same rational map have different Jacobians as matrices over $k^{N \times n}$. So $d\varphi_p$ is not determined as a map on the ambient $k^n$. The claim is that, despite this, the restriction of the Jacobian to the subspace $T_{V, p}$ is unambiguous. The argument rests on two facts used in parallel. \textbf{Fact 1: Polynomials agreeing on a dense open subset are equal modulo $I(V)$.} Two polynomial presentations $f_j$ and $\tilde f_j$ of the same rational function agree as functions on their common domain $U \cap \tilde U$, which is a nonempty open subset of the irreducible variety $V$. By irreducibility, $U \cap \tilde U$ is dense in $V$; a polynomial that vanishes on a dense subset of $V$ vanishes on all of $V$ as a regular function, i.e. lies in $I(V)$. So $f_j - \tilde f_j \in I(V)$. \textbf{Fact 2: Gradients at $p$ of elements of $I(V)$ kill $T_{V, p}$.} This is literally the definition of $T_{V, p}$: $T_{V, p} = \{v \in k^n : \nabla g(p) \cdot v = 0 \text{ for all } g \in I(V)\}$. Combining: applied to $g = f_j - \tilde f_j \in I(V)$, Fact 2 gives $\nabla(f_j - \tilde f_j)(p) \cdot v = 0$ for $v \in T_{V, p}$, so the $j$-th rows of the two Jacobians agree on $T_{V, p}$. Why does the same two-step pattern show $d\varphi_p$ lands in $T_{W, \varphi(p)}$? The setup is parallel: for any $h \in I(W)$, $h \circ \varphi$ vanishes on $V$ (because $\varphi(U) \subset W$ and Fact 1), so $h \circ \varphi \in I(V)$, so $\nabla(h \circ \varphi)(p) \cdot v = 0$ for $v \in T_{V, p}$ (Fact 2). The chain rule $\nabla(h \circ \varphi)(p) = \nabla h(\varphi(p)) \cdot J\varphi(p)$ converts this into $\nabla h(\varphi(p)) \cdot (J\varphi(p) \cdot v) = 0$, so $J\varphi(p) \cdot v$ is annihilated by every gradient $\nabla h(\varphi(p))$ with $h \in I(W)$, hence lies in $T_{W, \varphi(p)}$.[/guided]

[step:Establish the chain rule] Let $\psi : W \dashrightarrow Z \subset \mathbb{A}^M_k$ be a rational map defined at $\varphi(p)$, given locally by $\psi = (h_1, \ldots, h_M)$ with $h_\ell \in k[Y_1, \ldots, Y_N]$ (or rational in $Y$ with denominators nonvanishing at $\varphi(p)$). On a neighbourhood of $p$ where both are defined, \begin{align*} (\psi \circ \varphi)(x) = (h_1(f_1(x), \ldots, f_N(x)), \ldots, h_M(f_1(x), \ldots, f_N(x))). \end{align*} The multivariate chain rule (a purely algebraic identity for polynomials in several variables) gives \begin{align*} \frac{\partial (h_\ell \circ \varphi)}{\partial X_i}(p) = \sum_{j=1}^{N} \frac{\partial h_\ell}{\partial Y_j}(\varphi(p)) \cdot \frac{\partial f_j}{\partial X_i}(p), \end{align*} i.e. as matrices, \begin{align*} J(\psi \circ \varphi)(p) = J\psi(\varphi(p)) \cdot J\varphi(p) \in k^{M \times n}. \end{align*} For $v \in T_{V, p}$: \begin{align*} d(\psi \circ \varphi)_p(v) = J(\psi \circ \varphi)(p) \cdot v = J\psi(\varphi(p)) \cdot (J\varphi(p) \cdot v) = d\psi_{\varphi(p)}(d\varphi_p(v)). \end{align*} The last equality uses $d\varphi_p(v) \in T_{W, \varphi(p)}$ from Step 2, so $d\psi_{\varphi(p)}$ is validly applied (and lands in $T_{Z, \psi(\varphi(p))}$ by the same Step-2 argument applied to $\psi$). Therefore \begin{align*} d(\psi \circ \varphi)_p = d\psi_{\varphi(p)} \circ d\varphi_p : T_{V, p} \to T_{Z, (\psi \circ \varphi)(p)}, \end{align*} which is the chain rule. [/step]