= Solution
The <Uniform boundedness principle> states that a pointwise bounded family $\mathcal T$ of bounded <linear maps> from a <Banach space> $X$ to a normed space $Y$ is uniformly bounded in <operator norm>. Define closed sets $E_n=\{x:\sup_{T\in\mathcal T}\|Tx\|\le n\}$. Pointwise boundedness makes their union $X$. The <Baire category theorem> gives a ball $B(x_0,r)\subset E_n$ for some $n$. For $\|v\|<r$, both $x_0+v$ and $x_0$ lie in $E_n$, so $\|Tv\|\le2n$ for every $T$. Scaling yields $\sup_T\|T\|\le2n/r$, proving the principle.
For the separately continuous <bilinear map>, consider $T_x(y)=F(x,y)$ with $\|x\|\le1$. Each $T_x$ is bounded and linear. For fixed $y$, continuity of the other <linear map> gives $\sup_{\|x\|\le1}\|F(x,y)\|<\infty$. Apply the <Uniform boundedness principle> to this family to get
$$
\boxed{\|F(x,y)\|\le M\|x\|\|y\|.}
$$
Bilinearity then gives $F(x,y)-F(x_0,y_0)=F(x-x_0,y)+F(x_0,y-y_0)$; the bound proves joint continuity.
For the trilinear map, fix $w$. The just-proved bilinear result bounds $G(\cdot,\cdot,w)$. Thus the bounded <linear maps> $T_{x,y}:w\mapsto G(x,y,w)$, for $\|x\|,\|y\|\le1$, are pointwise bounded. Apply the <Uniform boundedness principle> once more to obtain $\|G(x,y,w)\|\le C\|x\|\|y\|\|w\|$. Expansion of the difference in three slots proves joint continuity. \b[Separate continuity suffices for these multilinear maps.]
Without linearity it does not suffice. On $\mathbb R^2$ set $H(x,y)=xy/(x^2+y^2)$ away from the origin and $H(0,0)=0$. Each coordinate section is continuous, but $H(t,t)=1/2$ for $t\ne0$, so $H$ is not continuous at the origin.
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