We are given a function,$$f(x,y)=4x^2-xy+4y^2+x^3y+xy^3-4$$ Now to calculate minimum and maximum value of $f(x,y)$, i first calculated $f_x$ and $f_y$ for stationary points. which gave $$f_x=8x-y+3x^2y+y^3---(equation\ a) $$ and $$f_y=-x+8y+x^3+3xy^2---(equation\ b)$$ $f_x=0\ and\ f_y=0$ gives the stationary point but i am unable to find it. I tried substitution method but could not get a valid substitute, i also tried
$y*equation(a) - x*equation(b)$ which gave me $x^2-y^2+y^4-x^4=0,$ but it does not yield (1.5,-1.5) and (-1.5,1.5) which I obtained from graph,
how can i obtain intersection point without graphing?
If you make $y=x$ you would get $$f_x=x \left(4 x^2+7\right)=0\implies x=0 \implies y= ???$$ If you make $y=-x$ you would get $$f_x=9 x-4 x^3=0\implies x=\pm \frac 32\implies y= ???$$
Then, all the solutions.