The vector pointing from (2, 1) to (1, 3) points in the same direction as the vector [tex]\vec u=(1,3)-(2,1)=(-1,2)[/tex]. The derivative of [tex]f[/tex] at (2, 1) in the direction of [tex]\vec u[/tex] is
[tex]D_{\vec u}f(2,1)=\nabla f(2,1)\cdot\dfrac{\vec u}{\|\vec u\|}[/tex]
We have
[tex]\|\vec u\|=\sqrt{(-1)^2+2^2}=\sqrt5[/tex]
Then
[tex]D_{\vec u}f(2,1)=(f_x(2,1),f_y(2,1))\cdot\dfrac{(-1,2)}{\sqrt5}=\dfrac{-f_x(2,1)+2f_y(2,1)}{\sqrt5}=-\dfrac2{\sqrt5}[/tex]
[tex]\implies f_x(2,1)-2f_y(2,1)=2[/tex]
The vector pointing from (2, 1) to (5, 5) has the same direction as the vector [tex]\vec v=(5,5)-(2,1)=(3,4)[/tex]. The derivative of [tex]f[/tex] at (2, 1) in the direction of [tex]\vec v[/tex] is
[tex]D_{\vec v}f(2,1)=\nabla f(2,1)\cdot\dfrac{\vec v}{\|\vec v\|}[/tex]
[tex]\|\vec v\|=\sqrt{3^2+4^2}=5[/tex]
so that
[tex](f_x(2,1),f_y(2,1))\cdot\dfrac{(3,4)}5=1[/tex]
[tex]\implies3f_x(2,1)+4f_y(2,1)=5[/tex]
Solving the remaining system gives [tex]f_x(2,1)=\dfrac95[/tex] and [tex]f_y(2,1)=-\dfrac1{10}[/tex].
