Rewriting some equations of your problem for convenience
$$h(x) = h_1(x), \, h(y) = h_2(y)$$
$$g(x,y) = \left[\begin{matrix}g_1(x,y)\\g_2(x,y)\end{matrix}\right]$$
$$f(h(x),g(x,y),h(y)) = f(h_1(x),g_1(x,y),g_2(x,y),h_2(y)) = \left[\begin{matrix}f_1(h_1(x),g_1(x,y),g_2(x,y),h_2(y))\\f_2(h_1(x),g_1(x,y),g_2(x,y),h_2(y))\end{matrix}\right] = \left[\begin{matrix}0\\0\end{matrix}\right]$$
$$J_f(h_1(x),g_1(x,y),g_2(x,y),h_2(y)) = \left[\matrix{\frac{\partial f_1}{\partial h_1}\\\frac{\partial f_2}{\partial h_1}}\matrix{\frac{\partial f_1}{\partial g_1}\\\frac{\partial f_2}{\partial g_1}}\matrix{\frac{\partial f_1}{\partial g_2}\\\frac{\partial f_2}{\partial g_2}}\matrix{\frac{\partial f_1}{\partial h_2}\\\frac{\partial f_2}{\partial h_2}}\right]$$
$$J_g(x,y) = \left[\matrix{\frac{\partial g_1}{\partial x}\\\frac{\partial g_2}{\partial x}}\matrix{\frac{\partial g_1}{\partial y}\\\frac{\partial g_2}{\partial y}}\right]$$
$$J_{h_1}(x) = \frac{\partial h_1}{\partial x}$$
$$J_{h_2}(y) = \frac{\partial h_2}{\partial y}$$
From the chain rule, we know that for $i={1,2}$
$$ \frac{\partial f_i}{\partial x} = 0 = \frac{\partial f_i}{\partial h_1} \frac{\partial h_1}{\partial x} + \frac{\partial f_i}{\partial g_1} \frac{\partial g_1}{\partial x} + \frac{\partial f_i}{\partial g_2} \frac{\partial g_2}{\partial x} $$
$$ \frac{\partial f_i}{\partial y} = 0 = \frac{\partial f_i}{\partial h_2} \frac{\partial h_2}{\partial y} + \frac{\partial f_i}{\partial g_1} \frac{\partial g_1}{\partial y} + \frac{\partial f_i}{\partial g_2} \frac{\partial g_2}{\partial y} $$
It is given that
$J_f(h(x_0),g_1(x_0,y_0),g_2(x_0,y_0),h(y_0)) = \bigg[\matrix{1\\-2}\matrix{7\\0}\matrix{0\\-5}\matrix{2\\5}\bigg], J_{h_1}(x_0) = \big[\matrix{14}\big], J_{h_2}(y_0) = \big[\matrix{10}\big]$
So it is possible to write the system
$$ 14 + 7 \frac{\partial g_1}{\partial x} (x_0,y_0) = 0 $$
$$ 20 + 7 \frac{\partial g_1}{\partial y} (x_0,y_0) = 0 $$
$$ -28 -5 \frac{\partial g_2}{\partial x} (x_0,y_0) = 0 $$
$$ 50 - 5 \frac{\partial g_2}{\partial y} (x_0,y_0) = 0 $$
Best Answer
Let $f:ℝ^n→ℝ^m$ be a function with arguments $(x_1,…,x_n)$ and components $(f_1,…f_m)$. The Jacobian is defined as the $m×n$ system: $$J(f) = \left(\frac{∂f_i}{∂x_j}\right)_{i,j=1}^{m,n}=\pmatrix{\frac{∂f_1}{∂x_1} & \frac{∂f_1}{∂x_2} & \dots & \frac{∂f_1}{∂x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{∂f_m}{∂x_1}& \frac{∂f_m}{∂x_2} & \dots &\frac{∂f_m}{∂x_n}} $$
In your case it is $$f(x, y, z) = \pmatrix{x^2 + \ln(y) \\ \sqrt{y − z}}=: \pmatrix{f_1(x,y,z)\\f_2(x,y,z)}. $$
So the Jacobian is a $2×3$ matrix
$$J(f) = \pmatrix{2x& \star & \star\\ 0 & \star & \star}$$
Can you take it from here?