= Solution
In local <holomorphic coordinates>, write the two positive <singular values> of $Df$ as $s_1\geq s_2$. Preservation of <orientation> gives $J_f=s_1s_2$, and <quasiconformality> gives $s_1/s_2\leq K$.
For an admissible <conformal metric> $\rho|dw|$ on $Y$, define a density on $X$ by $\sigma(z)=\rho(f(z))s_1(z)$. Along each path,
$$
\int_{f(\gamma)}\rho|dw|
\leq\int_\gamma\sigma|dz|,
$$
so $L_\rho(f(\Gamma))\leq L_\sigma(\Gamma)$. The <change of variables formula> and $s_1^2\leq KJ_f$ give
$$
A_\sigma=\int_X\rho(f(z))^2s_1(z)^2\,dx\,dy
\leq K\int_X\rho(f(z))^2J_f(z)\,dx\,dy
=KA_\rho.
$$
Also $s_1^2\geq J_f$, so $A_\sigma\geq A_\rho>0$. Thus $\sigma$ has positive finite area and is admissible, and
$$
\frac{L_\rho(f(\Gamma))^2}{A_\rho}
\leq K\,\frac{L_\sigma(\Gamma)^2}{A_\sigma}
\leq K\,\lambda(\Gamma,X).
$$
Taking the <supremum> over $\rho$ proves
$$
\boxed{\lambda(f(\Gamma),Y)\leq K\lambda(\Gamma,X).}
$$
Applying the same argument to $f^{-1}$, which has the same bound on its <maximal dilatation>, also gives $K^{-1}\lambda(\Gamma,X)\leq\lambda(f(\Gamma),Y)$. The inequalities remain valid for extended <extremal lengths>; no extremizing density need exist.
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