Find the curvature of the curve defined by: $$C:=\begin{cases} \left(x+y\right)^{2}-z=yz \\ \left(x-z\right)^{2}=y+z-1\\ \end{cases}$$ At the point $(1,0,1)$.
All of the curves for which I found them their curvature at the given point were some vector-valued functions,but this one is different and I don't know how to find the curvature.
On
The curve is $$C:X(z)=\{x(z),y(z),z\},\tag 1$$ where $$(x(z)+y(z))^2=y(z)z+z\tag 2$$ and $$(x(z)-z)^2=y(z)+z-1.\tag 3$$ We can see that indeed $P=(1,0,1)$ is a point on the curve i.e. $x(1)=1$, $y(1)=0$. Differentiating (2),(3) we get $$2(x(z)+y(z))(x'(z)+y'(z))=y'(z)z+y(z)+1\tag 4$$ and $$2(x(z)-z)(x'(z)-1)=y'(z)+1.\tag 5$$ Hence when $x(1)=1$, $y(1)=0$, we $x'(1)=1$, $y'(1)=-1$. Hence $$\dot{X}(1)=\{1,-1,1\}.$$ Differentiating again (4) and (5), we get $$2(x'(z)+y'(z))^2+2(x''(z)+y''(z))(x(z)+y(z))=2y'(z)+zy''(z)\tag 6$$ and $$2(-1+x'(z))^2+2(-z+x(z))x''(z)=y''(z).\tag 7$$ Setting $x(1)=1$, $y(1)=0$ and $x'(1)=1$, $y'(1)=-1$, we get $x''(1)=-1$, $y''(1)=0$. Hence $$\ddot{X}(1)=\{-1,0,0\}.$$ The curvature in any point of the curve is $$ k=\frac{\left\|\dot{X}\times \ddot{X}\right\|}{\left\|\dot{X}\right\|^3}\tag 8 $$ But $$(\dot{X}\times \ddot{X})_{P}=\{0,-1,-1\}.$$ Also easily $$\left\|\dot{X}\right\|^2=(\dot{X}\cdot\dot{X})_P=3$$ Hence the curvature is $$k_P=\frac{1}{3}\sqrt{\frac{2}{3}}\tag 9$$
Unless there is some trick specific to curve you're asking, actually calculating the curvature is quite tedious so I am not going to do this here. Instead, I'll guide you through the "general" method of calculating it.
Our first task is to find the direction of the velocity vector at that point. The intuition is that the gradient of a function is perpendicular to its level set so the velocity of a curve at the intersection of two level sets has to be perpendicular to both gradients. We conclude it must lie in the direction of their cross product which is perpendicular to both. More explicitly: start by taking a parametrization by arc length $c$ of your curve around $(1, 0, 1)$, that is, let's say $$f(x, y, z) = (x + y)^2 - z - yz$$ $$g(x,y,z) = (x-z)^2 - y - z + 1$$ then $$c(t): (-\epsilon, \epsilon)\rightarrow\mathbb{R}^3$$ satisfies $$c(0) = (1, 0, 1)$$$$f(c(t)) = g(c(t)) = 0$$ $$\tag{1}\|c'(t)\| = 1$$ for all $t \in (-\epsilon, \epsilon)$. Then we know that $$\tag{2} \frac{d}{dt}(f\circ c)(t) = \nabla f(c(t)) \cdot c'(t) = 0$$ and similarly $$\tag{3}\nabla g(c(t)) \cdot c'(t) = 0$$ There's only two possible vectors $c'(0)$ satifying constraints $(1)$, $(2)$ and $(3)$: $$c'(0) = \pm \frac{\nabla f(c(0)) \times \nabla g(c(0))}{\|\nabla f(c(0)) \times \nabla g(c(0))\|}$$ up to the direction of parametrization. This is geometrically intuitive but you can prove it by writing down the coordinates in equations $(1)$, $(2)$ and $(3)$ or replacing this value in those equations.
Now to know what $c''(0)$ is, we can derive equation $(2)$ and take its value at $0$ to get $$\tag{4} \mathbf{H} f_{c(0)}c'(0) \cdot c'(0) + \nabla f(c(0))\cdot c''(0) = 0$$ where $\mathbf{H}f_v$ is the Hessian of $f$ at v and similarly $$\tag{5} \mathbf{H} g_{c(0)}c'(0) \cdot c'(0) + \nabla g(c(0))\cdot c''(0) = 0$$ Alternatively, the left hand side of $(2)$ and $(3)$ are one variable real valued functions that you can derive directly by replacing with coordinates. Besides, the curvature vector has to be perpendicular to the velocity vector because $c$ is parametrized by arc length: $$\tag{6} c'(0) \cdot c''(0) = 0 $$ Equations $(4)$, $(5)$ and $(6)$ are linear equations in the components of $c''(0)$ which should allow it to be calculated directly.