Show that the function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ given by
$$f(x,y)=x^4+6x^2y^2 + y^4 -\frac{9}{4}x-\frac{7}{4}$$
achieves its minimal value, and determine all the points $(x, y)\in\mathbb{R}^2$ at which it is achieved.
This function is continuous on $\mathbb{R}^2$ and $\displaystyle\lim_{(x, y)\to\infty}f(x,y)=\infty$. Therefore it must achieve its minimal at some point $(x, y)$. At this point, it must hold that
$$\frac{\partial{f}}{\partial{x}}=\frac{\partial{f}}{\partial{y}}=0$$
which means
$$\left\{\begin{array}{rl} 4x^3 + 12xy^2 -\frac{9}{4}&=0\\ 12x^2y + 4y^3 -\frac{7}{4}&=0\end{array}\right.$$
or equivalently
$$\left\{\begin{array}{rl} x^3 + 3xy^2 &=\frac{9}{16}\\ 3x^2y + y^3 &=\frac{7}{16}\end{array}\right.$$
The sum of the left sides of these two equations equals a perfect cube $(x+y)^3$. Hence, we find
$$(x+y)^3=1 \implies x+y=1$$
Similarly, the difference of their left sides is also a perfect cube $(x-y)^3$. Hence, we have
$$(x-y)^3=\frac{1}{8}\implies x-y=\frac{1}{2}$$
Therefore, we find $(x, y)=\boxed{\left(\frac{3}{4},\frac{1}{4}\right)}$