= Solution
At $y=0$, both fractions equal $x=0.0402$. Reading the original plot near this value gives representative <momentum> densities
$$
U=xu_v\simeq0.42,\qquad D=xd_v\simeq0.24,\qquad 0.05x\Sigma\simeq0.08.
$$
These are approximate graphical readings, not a precision <Parton distribution function>[PDF] fit. Undo the sea-curve scale and divide by eight <Flavor eigenstates>[flavours]/antiflavours:
$$
S_x=xS=\frac{x\Sigma}{8}\simeq\frac{0.08}{0.05\times8}=0.20.
$$
All products have the same $1/x^2$ factor, which cancels in the fractions. The three relative weights are
$$
w_{VV}=UD=0.1008,\qquad w_{VS}=S_x(U+D)=0.1320,\qquad w_{SS}=2S_x^2=0.0800.
$$
Their sum is $0.3128$, giving
$$
\boxed{f_{VV}\simeq32\%,\qquad f_{VS}\simeq42\%,\qquad f_{SS}\simeq26\%.}
$$
Reading the central curves more finely changes these percentages by about a percentage point, well within what the plot supports. Omitting its factor $0.05$ would substantially underestimate the sea contribution.
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