The answer is #f(theta)=e^(theta)# (and #f(theta)=e^(-theta)#, depending on how you decide to measure the angle, see below).
You can think about this in terms of dot products of appropriate vectors in rectangular coordinates. Let #r=f(theta)# be the unknown polar function you want to find. Then #x=f(theta)cos(theta)# and #y=f(theta)sin(theta)#. As #theta# varies, this corresponds to a varying position vector for the point #P# that is #vec(r)=f(theta)cos(theta)hat(i)+f(theta)sin(theta)hat(j)#.
The velocity vector #vec(v)=vec(r)'# is tangent to the resulting parametric curve (where the angular coordinate #theta# is the parameter). Therefore, the dot product #vec(r)\cdot vec(v)=|| vec(r) || || vec(v) || cos(alpha)#, where you want #alpha# to be the constant angle between #vec(r)# and #vec(v)#: #alpha=pi/4=45^{\circ}# so that #cos(alpha)=1/sqrt(2)# and #vec(r)\cdot vec(v)=1/sqrt(2)|| vec(r) || || vec(v) ||#.
Now, by the Product Rule,
#vec(r)\cdot vec(v)=(f(theta)cos(theta)hat(i)+f(theta)sin(theta)hat(j))\cdot ((f'(theta)cos(theta)-f(theta)sin(theta))hat(i)+(f'(theta)sin(theta)+f(theta)cos(theta))hat(j))#.
Through expansion of this dot product and the use of the trigonometric identity #cos^{2}(theta)+sin^{2}(theta)=1#, this reduces to #vec(r)\cdot vec(v)=f(theta)f'(theta)#. Moreover, the same standard trigonometric identity can be used to derive the fact that, assuming #f(theta)\geq 0#, we have #|| vec(r) || || vec(v) ||=f(theta)\sqrt((f(theta))^2+(f'(theta))^2)#. (Check these claims!)
The condition #vec(r)\cdot vec(v)=1/sqrt(2)|| vec(r) || || vec(v) ||# thus reduces to, after cancelation of #f(theta)# (assuming #f(theta)>0#), to #f'(theta)=1/sqrt(2)\sqrt((f(theta))^2+(f'(theta))^2)#. Squaring both sides, rearranging, and multiplying both sides by 2 ultimately gives #(f'(theta))^2=(f(theta))^2#. This is really a differential equation for #f(theta)#, which in Leibniz notation would be written as #(\frac{dy}{d\theta})^2=y^2#. Certainly #y=f(theta)=e^{theta)# is one solution of this equation that satisfies #y(0)=1#. The function #y=f(theta)=e^{-theta}# also satisfies the given initial value problem.
Is it possible one of these solutions is extraneous (not a solution of the original problem)? It's certainly possible since we've squared both sides of an equation in our work.
The answer, in a sense, depends on how you measure the angle #alpha#. If it is measured as the angle between the vectors #vec(r)# and #vec(v)# in the standard way (put their base points at the same location), then the second answer turns out to be extraneous.
I'll assume the desired angle is the angle between the two vectors. Based on this assumption, the unique answer is #f(theta)=e^{theta}#.
