= Solution
Continuity on the compact product of unit spheres bounds the <bilinear map>: there is $C$ such that $\|f(h,k)\|_3\le C\|h\|_1\|k\|_2$. This follows by scaling each nonzero vector to unit <norm>; if one factor space is zero, the map and the relevant assertion are trivial.
Bilinearity gives the exact expansion
$$
f(a+h,b+k)-f(a,b)=f(h,b)+f(a,k)+f(h,k).
$$
The proposed <derivative> $L(h,k)=f(h,b)+f(a,k)$ is linear and bounded, with $\|L(h,k)\|_3\le C(\|b\|_2\|h\|_1+\|a\|_1\|k\|_2)$. Set $\rho=(\|h\|_1^2+\|k\|_2^2)^{1/2}$. Its remainder obeys
$$
\frac{\|f(h,k)\|_3}{\rho}\le C\frac{\|h\|_1\|k\|_2}{\rho}\le\frac C2\rho\longrightarrow0.
$$
Thus the <derivative of a continuous bilinear map> is the <Fréchet derivative>
$$
\boxed{Df(a,b)(h,k)=f(h,b)+f(a,k).}
$$
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