This probably can be done with some clever manipulation of inequalities but the problem does suggest that the method of Lagrange multiplier would be effective.
If we define
$F = \dfrac{a^3}{c} + \dfrac{b^3}{d} + \lambda\left(c^2+d^2-(a^2+b^2)^3 \right)$,
so
\begin{eqnarray}
F_a &=& 3\,\frac {a^2}{c}-6a\,\lambda \left( a^2+b^{2} \right) ^{2}\\
F_b &=&3\,\frac {b^2}{d}-6b\,\lambda \left( a^2+b^{2} \right) ^{2}\\
F_c &=&-\,\frac {a^3}{c^2}+2\,\lambda c\;\;\;\;\;(*)\\
F_d &=&-\,\frac {b^3}{d^2}+2\,\lambda d\\
F_{\lambda} &=&c^2+d^2-(a^2+b^2)^3
\end{eqnarray}
Setting $F_c=0$ and solving for $\lambda$ gives $\lambda= \dfrac {a^3}{2c^3}$. From the $F_a=0$ equation we have
$-3\,{\dfrac {{a}^{2} \left( -c+{a}^{3}+a{b}^{2} \right) \left( c+{a}^{
3}+a{b}^{2} \right) }{{c}^{3}}}
=0$
from which we obtain $c = a^3+ab^2$. Factoring the $F_b=0$ equation gives
$3\,{\dfrac { \left( {a}^{2}b+{b}^{3}-d \right) b}{ \left( {a}^{2}+{b}^{
2} \right) d}}=0$
leading to $d = a^2b+b^3$ noting that the remaining equation in (*) are satisfied. Thus, we have
$\dfrac{a^3}{c} + \dfrac{b^3}{d} = \dfrac{a^3}{a^3+ab^2} + \dfrac{b^3}{a^2b+b^3} = 1$
the minimum as required.