Given $f(x,y)$ differentiable at $(a,b)$ and $D_{\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)} f(a,b)=3$, $D_{\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)}f(a,b)=1$, find $f_x(a,b)$ and $f_y(a,b)$.
I know $D_af= \mathrm{grad}(f) ⋅ a = a_1⋅f_x + a_2⋅f_y$.
In that case, I'd have \begin{equation} D_{\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)}f(a,b)= \frac{1}{\sqrt{2}}f_x(a,b) + \frac{1}{\sqrt{2}}f_y(a,b) \end{equation} I have tried to go about this by breaking up the equation, but don't know how to incorporate 3 and 1 to find the gradient.
As you say \begin{equation} D_{\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)}f(a,b)= \frac{1}{\sqrt{2}}f_x(a,b) + \frac{1}{\sqrt{2}}f_y(a,b). \end{equation} The left hand side is equal to 3 by assumption. Similarly \begin{equation} 1 = D_{\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)}f(a,b)= \frac{1}{\sqrt{2}}f_x(a,b) - \frac{1}{\sqrt{2}}f_y(a,b). \end{equation} You get a linear system where the unknowns are exactly $f_x(a,b)$ and $f_y(a,b)$.