The problem is find to the differential of $f:S^{m}\times S^{n}\to S^{m+n+mn}$ (spheres) defined as $f(x_{0},...,x_{n},y_{1},...y_{n})=(x_{0}y_{0},x_{0}y_{1},...,x_{n}y_{1},x_{n}y_{n})$ and show it is injective for all $(p,q)\in S^{m}\times S^{n}$ i.e. $df_{(p,q)}:T_{p}S^{m}\times T_{q}S^{n}\to T_{f(p,q)}S^{m+n+mn}$ is injective.
one way) Treating f as a (m +1)x(n+1) matrix with $f_{ij}=x_{i}y_{j}$. But differentiating such a function is confusing.
other way) using coordinate charts. But there must be cleaner solution.
My suggestion is to avoid coordinates on the spheres as much as possible. Here's a simpler set-up. Suppose you have a smooth map $F\colon \Bbb R^{n+1}\to \Bbb R^N$ (the issue here is in the domain, not the range) and you're interested in its restriction $f\colon S^n\to\Bbb R^N$. How should we decide if $df_x\colon T_xS^n\to\Bbb R^N$ is injective? Of course, if $dF_x\colon\Bbb R^{n+1}\to\Bbb R^N$ is injective, we're very happy. But it needn't be. So remember that $df_x$ is the restriction of $dF_x$ to $T_xS^n$. Thus, it will suffice for $\ker(dF_x)$ to be contained in $(T_xS^n)^\perp=\text{Span}(x)$. You might compare this with the way the method of Lagrange multipliers works, by the way.
You should be able to generalize this to the case of $S^m\times S^n$.