You should prove that the tangent vector of the curve at $(1,1,1)$ is contained in the tangent plane of the surface, or is orthogonal to the normal vector to the surface at $(1,1,1)$ which is given by the gradient. You calculated the normal vector which is $(-1,-5,1)$.
Now you need to calculate the tangent vector to the curve in $(1,1,1)$. The curve is given by the intersection of two surfaces, so the tangent vector would be orthogonal to both normal vectors in $(1,1,1)$ of the two surfaces.
$(S_1): x^2-y^2+z^2=1$ has normal vector $(2x,-2y,2z)$ which evaluated in your point gives $v_2=(2,-2,2)$.
$(S_2): xy+xz=2$ has normal vector $(y+z,x,x)$ which evaluated in your point gives $v_3=(2,1,1)$.
Solve now the system given by $v\cdot v_2=0$ and $v\cdot v_3=0$ to get $v_1-v_2+v_3=0$ and $2v_1+v_2+v_3=0$. Eliminate $v_1$ to get $3v_2-v_3=0$. So if $v_2=\alpha$ we have $v_3=3\alpha$ and $v_1=-2\alpha$. So $v=\alpha(-2,1,3)$.
The vector $v$ is the tangent vector to the curve at $(1,1,1)$, and it should be orthogonal to $(-1,-5,1)$. Indeed if we calculate the scalar product we get $(-2,1,3)(-1,-5,1)=2-5+3=0$.
Best Answer
They want "a curve" - because there are an infinite number of curves that will satisfy the conditions!
So you need $y=7$ when $x=5$ and $\frac {dy}{dx}=2$ when $x=5$.
The simplest kind of curve would be a quadratic; I would suggest $y=x^2+ax+b$.
Substitute the known conditions into that curve to create simultaneous equations for $a$ and $b$.
Solve and there you are!