Let $X$ and $Y$ be smooth varieties, then $X\times Y$ is a smooth variety.
Here is my definition of smooth: 
It is enough to show that if $X$ is smooth at $p$ and $Y$ is smooth at $q$ then the product is smooth at $(p,q)$. The smoothness of $X$ at $P$ gives us a $U,\phi,(f_1,..,f_a)$ where $a=n-d$ with the Jacobi having maximal rank. Simillary the smoothness of $Y$ gives a $V,\psi,(g_1,…g_b)$ with $b=m-d'$ such that the Jacobi has maximal rank.
By definition the the product $U\times V$ is open in $X\times Y$ and is also a variety.
I think $\phi\times\psi$ should give an isomorphism to $Z(f_1,..,f_a,g_1,..,g_b)$. But this is not obvious to me. It is also not clear why the jacobi of this should be of maximal rank.
Best Answer
$\phi\times\psi$ does indeed give an isomorphism of $U\times V$ with $Z(f_1,\cdots,f_a)\times Z(g_1,\cdots,g_b) \subset \Bbb A^n\times\Bbb A^m$, which is exactly $Z(f_1,\ldots,f_a,g_1,\ldots,g_b)\subset \Bbb A^{n+m}$. We can check the first part by noting that $\phi^{-1}\times\psi^{-1}$ is an inverse to $\phi\times\psi$, and the second part is true because $(x_1,\ldots,x_n)$ satisfies all the $f_i$ and $(y_1,\ldots,y_m)$ satisfies all the $g_j$ iff $(x_1,\ldots,x_n,y_1,\ldots,y_m)$ satisfies all the $f_i$ and the $g_j$.
To check smoothness, it's instructive to write down the Jacobian matrix:
$$\begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \ldots & \frac{\partial f_1}{x_n} & \frac{\partial f_1}{\partial y_1} & \ldots & \frac{\partial f_1}{\partial y_m} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_a}{\partial x_1} & \ldots & \frac{\partial f_a}{x_n} & \frac{\partial f_a}{\partial y_1} & \ldots & \frac{\partial f_a}{\partial y_m} \\ \frac{\partial g_1}{\partial x_1} & \ldots & \frac{\partial g_1}{x_n} & \frac{\partial g_1}{\partial y_1} & \ldots & \frac{\partial g_1}{\partial y_m} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ \frac{\partial g_b}{\partial x_1} & \ldots & \frac{\partial g_b}{x_n} & \frac{\partial g_b}{\partial y_1} & \ldots & \frac{\partial g_b}{\partial y_m} \\ \end{pmatrix}$$
but $\frac{\partial f_i}{y_j}=0$ and $\frac{\partial g_i}{x_j}=0$ because no $f$ contains any $y$ and no $g$ contains any $x$. So our Jacobian matrix is of the form $$\begin{pmatrix} J_{f,x} & 0 \\ 0 & J_{g,y} \end{pmatrix}$$ where $J_{f,x}$ is the Jacobian of the $f$s with respect to the $x$s and similarly with $J_{g,y}$. But this exactly means that the big Jacobian is of maximal rank since the small ones are.