Showing $(L \leftrightarrow ((N \to N) \to L)) \lor H$ is not contingent.

Working P.D. Magnus. forallX: an Introduction to Formal Logic (p. 182, exercise B. 4). To solve it, I tried that sentence is a theorem:

$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}} \def\Ae#1{\qquad\mathbf{\forall E} \: #1 \\} \def\R#1{\qquad\mathbf{R} \: #1 \\} \def\ii#1{\qquad\mathbf{\to I} \: #1 \\} \def\ne#1{\qquad\mathbf{\neg E} \: #1 \\} \def\Ee#1{\qquad\mathbf{\exists E} \: #1 \\} \def\IP#1{\qquad\mathbf{IP} \: #1 \\} \def\ei#1{\qquad\mathbf{\exists I} \: #1 \\} \def\ie#1{\qquad\mathbf{\to E} \: #1 \\} \def\bi#1{\qquad\mathbf{\leftrightarrow I} \: #1 \\} \def\oi#1{\qquad\mathbf{\lor I} \: #1 \\}$$ $$\fitch{\, }{ \fitch{1.\, L}{ \fitch{2.\, N \to N}{ 3.\, L \R{1} }\\ 4.\, (N \to N) \to L \ii{2-3} }\\ \fitch{5.\, (N \to N) \to L}{ \fitch{6.\, N}{ 7.\, N \R{6} }\\ 8.\,N \to N \ii{6-7} 9.\, L \ie{11,5} }\\ 10.\,(L \leftrightarrow ((N \to N) \to L) \bi{1-4,5-9} 11.\, (L \leftrightarrow ((N \to N) \to L) \lor H \oi{10} }$$

Is the usage of $$\mathbf{\leftrightarrow I}$$ a good approach in this case?

It is evident (as you saw) that you'll need to prove the given wff is a tautology by proving the first disjunct. The proof you then give is the neatest and most direct option, establishing a biconditional by separately proving each direction by a Conditional Proof ($$\to$$I).