Changing to polar coordinates using the pass transformations

#{(x=r costheta),(y=rsin theta):}#

we have

#f(x,y)->f(r,theta)=r^2(7cos^2theta+4sintheta cos theta+3sin^2theta)#

and the condition

#x^2+y^2=1->r^2=1# then the transformed problem is

#{min,max}f(1,theta) = {min,max}(7cos^2theta+4sintheta cos theta+3sin^2theta)#

but after simplifications

#7cos^2theta+4sintheta cos theta+3sin^2theta=5+2(sin(2theta)+cos(2theta))#

so

#f(1,theta)=5+2(sin(2theta)+cos(2theta))#

The stationary points condition is

#(df)/(d theta) = 4(sin(2theta)+cos(2theta))=0# and this is attained at

#theta = 1/2(pm pi/4 + 2kpi)# for #k=0,1,2,cdots#

The stationary points qualification is done evaluating

#(d^2f)/(d theta^2) = -8(sin(2theta)+cos(2theta))# resulting in

#{min,max}={8 sqrt[2], -8 sqrt[2]}#

The values at those points #theta = {-pi/8, pi/8}# are

#m = 5 - 2 sqrt[2]# and #M = 5 + 2 sqrt[2]# so

#(m+M)/2=5#