Remind yourself what a partial derivative is. If we set $x$ as a constant, the partial derivative defines the slope of the curve as $y$ varies. Thus, if we take the partial derivative with respect to $y$ of the curve, we obtain the slope along the "slice" of $f(x,y)$, just as you require. Via partial differentiation, we acquire
$$\frac{\partial f}{\partial y}(2,y) = \frac{df}{dy}\left(2^2 + y^2\right) = 2y$$
since the $x^2$ is treated as constant. This is the formula for your slope.
Additionally, to find a vector tangent to a function at a point, we must consider both the partial derivatives. In this case, we calculate partials $\partial_x(x,y) = 2x$, $\partial_y(x,y) = 2y$. We then calculate the slope of the curve in 1. the $x$ direction and 2. the $y$ direction at point $(2,0,4)$; in this case, only the $(2,0)$ matters. We obtain that $\partial_x(2,0) = 4$ and $\partial_y(2,0) = 0$.
More simply, we may consider the gradient $\nabla f(x,y)$ which is a vector consisting of the two partial derivatives of $f$;
$$\nabla f(x,y) = \langle \partial_x(x,y), \partial_y(x,y) \rangle$$
It is unclear what your second problem means however, since there are infinitely many lines which we may say are tangent to $f(x,y)$. Thus, if clarification may be provided, I would be happy to finish this answer.
Best Answer
Since the plane is parallel to the $z$-axis, this question is basically asking you about a directional derivative of $f$ at the point $(2,-1)$. For this plane, $y$ is constant, so the particular directional derivative required is just the partial $x$-derivative of $f$ at this point ($\nabla f\cdot (1,0)=f_x$). I expect that you’ll be able to take it from here.