jmc

Cube both sides of $a+b=7$ to find $$a^3+3a^{2}b+3a{b}^2+b^3=343.$$ Substitute 42 for $a^3+b^3$ and factor $3ab$ out of the remaining two terms. $$\begin{array}{rl}42+3ab(a+b)& =343⟹\\ 3ab(a+b)& =301⟹\\ 3ab(7)& =301⟹\\ 3ab& =43⟹\\ ab& =\frac{43}{3}.\end{array}$$ Finally, $\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{7}{43/3}=\boxed{\frac{21}{43}}$.