Here is Prob. 6, Exercises 8.14, in the book Calculus Vol. II by Tom M. Apostol, 2nd edition:
Given a scalar field differentiable at a point $\mathbf{a}$ in $\mathbb{R}^2$, suppose that $f^\prime(\mathbf{a}; \mathbf{y})=1$ and $f^\prime(\mathbf{a}; \mathbf{z})=2$, where $\mathbf{y} = 2\mathbf{i}+3\mathbf{j}$ and $\mathbf{z} = \mathbf{i} + \mathbf{j}$. Make a sketch showing the set of all points $(x, y)$ for which $f^\prime(\mathbf{a}; x\mathbf{i}+y\mathbf{j})=6$. Also, calculate the gradient $\nabla f(\mathbf{a})$.
My Attempt:
As $f$ is differentiable at $\mathbf{a}$, so for every $\mathbf{b} \in \mathbb{R}^2$ the directional derivative $f^\prime(\mathbf{a}; \mathbf{b})$ exists and is given by $$ f^\prime(\mathbf{a}; \mathbf{b}) = \nabla f(\mathbf{a}) \cdot \mathbf{b}. $$
Now let $(u, v)$ be the gradient vector of $f$ at the point $\mathbf{a}$. Then from the given information we can conclude that $$ 2u + 3v = 1 \qquad \mbox{ and } \qquad u+v = 2, $$ and upon solving these two equations simultaneously for $u$ and $v$ we get $u = 5$ and $v=-3$. Thus $$ \nabla f(\mathbf{a}) = (5, -3). $$
Am I right?
Thus the set of all the points $(x, y)$ for which $f^\prime(\mathbf{a}; x\mathbf{i} + y\mathbf{j})=6$ is $$ \left\{ \ (x, y) \in \mathbb{R}^2 \ \colon \ 5x-3y=6 \ \right\}, $$ which is the straight line in the plane through the point (0, -2) and having the slope $5/3$.
Am I right?