maximum and minimum of $x^2+y^4$ for real $x,y$ If $y^2(y^2-6)+x^2-8x+24=0$ then maximum and minimum value of $x^2+y^4$ is 
what i try
$y^4-6y^2+9+x^2-8x+16=1$
$(x-4)^2+(y^2-3)^2=1\cdots (1)$
How i find maximum and minimum of $x^2+y^4$ from $(1)$ relation
help me to solve it please 

Use substitution method:
$$(x-4)^2+(y^2-3)^2=1 \Rightarrow y^2=\pm\sqrt{1-(x-4)^2}+3\\
f(x,y)=x^2+y^4 \Rightarrow \\
f(x)=x^2+1-(x-4)^2\pm6\sqrt{1-(x-4)^2}+9\\
f'(x)=8+\frac{\mp6(x-4)}{\sqrt{1-(x-4)^2}}=0 \Rightarrow \\
4\sqrt{1-(x-4)^2}=\pm3(x-4) \Rightarrow \\
16(-x^2+8x-15)=9(x^2-8x+16) \Rightarrow \\
25x^2-200x+384=0 \Rightarrow \\
x_1=\frac{16}{5}; x_2=\frac{24}{5}\\
f''\left(\frac{16}{5}\right)>0 \Rightarrow f\left(\frac{16}{5}\right)=16 \ \text{(min)}\\
f''\left(\frac{24}{5}\right)<0 \Rightarrow f\left(\frac{24}{5}\right)=36 \ \text{(max)}\\$$
