Line integral along the intersection of two surfaces Assume that the curve $L$ (in $\mathbb R^3$) is the intersection of two surfaces $F(x,y,z)=0$ and $G(x,y,z)=0$. Could we prove that the arc length of $L$ from $A(x_1,y_1,z_1)$ to $B(x_2,y_2,z_2)$ is
$$
  \int_{A}^{B}\mathrm ds=\int_{x_1}^{x_2}\sqrt{1+\Bigg(\frac{\partial(F,G)}{\partial(z,x)}\Bigg/\frac{\partial(F,G)}{\partial(y,z)}\Bigg)^2+\Bigg(\frac{\partial(F,G)}{\partial(x,y)}\Bigg/\frac{\partial(F,G)}{\partial(y,z)}\Bigg)^2}\mathrm dx\tag{1}
$$
where $A,B$ are two different points on $L$.
p.s. I tried to consider the arc length infinitesimal $\mathrm ds$ of $L$ since $\nabla F\times\nabla G$ is the tangent vector of $L$ and
$$
  \Vert\nabla F\times\nabla G\Vert_2=\sqrt{\Bigg(\frac{\partial(F,G)}{\partial(x,y)}\Bigg)^2+\Bigg(\frac{\partial(F,G)}{\partial(y,z)}\Bigg)^2+\Bigg(\frac{\partial(F,G)}{\partial(z,x)}\Bigg)^2}
$$
but I failed to relate this to integral on $x$ in $(1)$.
 A: Let $H:=(F,G)$ and assume that it is a $C^1$ function and that $(0,0)$ is a regular value. Then $\operatorname{rank}[\partial H(p)]=2$ at any $p=(x,y,z)\in L$ (where $L:=H^{-1}(0,0)$), so at least one of the $2\times 2$ submatrices of $[\partial H(p)]$ is not zero, so WLOG assume that for some fixed $p\in L$ we have that $\frac{\partial H(p)}{\partial (y,z)}\neq 0$.
As $\partial H$ is continuous then there is a neighborhood $U\subset \mathbb{R}^3$ of $p$ such that $\frac{\partial H(q)}{\partial (y,z)}\neq 0$ for all $q\in U$. Now observe that the map $\phi (q):=(q_1,H_1(q),H_2(q))=(q_1,F(q),G(q))$ for $q:=(q_1,q_2,q_3)$ is a diffeomorphism as it Jacobian doesn't vanish there. Also observe that $\phi(q)=(q_1,0,0)$ when $q\in U\cap L$, thus as $\phi$ is invertible in $U\cap L$ there is a function $g:U\to \mathbb{R}^2$ such that $H(p)=0$ in $U$ if and only if $p=(x,g_1(x),g_2(x))$ (this is just the implicit function theorem).
Therefore it follows that the map defined by $\tilde g(x):=(x,g_1(x),g_2(x))$ is a local parametrization of $L$. From the implicit function theorem we also knows that $[\partial g(x)]=-[D_2 H(x,g(x))]^{-1}[D_1 H(x,g(x))]$ where the brackets in $[D_2 f(h)]$ means that we are getting the matrix representation of the linear map $D_2 f(h)$, and the notation $D_j$ means derivative respect to the $j$-th argument (that can be a vector, by example $D_2H(x,g(x))$ means the derivative of $H$ respect to the vector $g(x)$).
In our case we have that for $v=(y,z)$
$$
\begin{align*}
&[D_2 H(x,v)]=[D_2(F(x,v),G(x,v))]=\begin{bmatrix}
\partial _2 F(x,y,z)&\partial _3F(x,y,z)\\
\partial _2G(x,y,z)&\partial _3G(x,y,z)
\end{bmatrix}\tag1\\[2em]
&[D_1 H(x,v)]=[\partial _1 F(x,y,z),\partial _1G(x,y,z)]\tag2\\[2em]
&\begin{bmatrix}
a&b\\c&d
\end{bmatrix}^{-1}=\frac{\begin{bmatrix}
d&-b\\-c&a
\end{bmatrix}}{\det \begin{bmatrix}
a&b\\c&d
\end{bmatrix}}\tag3
\end{align*}
$$
From the last three results above we finally get
$$
\tilde g'(t)=\left(1,\frac{\partial H (\tilde g(t))}{\partial (x,z)}/\frac{\partial H (\tilde g(t))}{\partial (y,z)},-\frac{\partial H (\tilde g(t))}{\partial (x,y)}/\frac{\partial H (\tilde g(t))}{\partial (y,z)}\right)\tag4
$$
And the length of $L$ between two any points $a,b\in U\cap L$ with $a_1<b_1$ is given by
$$
\begin{align*}
\int_{a_1}^{b_1}\|\tilde g'(t)\|_2 dt&=\int_{a_1}^{b_1}\sqrt{1+\left(\frac{\partial H (\tilde g(t))}{\partial (x,z)}/\frac{\partial H (\tilde g(t))}{\partial (y,z)}\right)^2+\left(\frac{\partial H (\tilde g(t))}{\partial (x,y)}/\frac{\partial H (\tilde g(t))}{\partial (y,z)}\right)^2}dt\\
&=\int_{a_1}^{b_1}\left|\frac{\partial H (\tilde g(t))}{\partial (y,z)}\right|^{-1}\|\nabla F(\tilde g(t))\times \nabla G(\tilde g(t))\|_2 dt\tag5
\end{align*}
$$
The notation is a bit cumbersome but needed to make an explicit formulation. Note that (5) is the same, but with a more clear notation, than what you had written in your question as the function $\tilde g$ make explicit the dependence of the coordinates $y$ and $z$ of $t$.
