How to prove a function has no local minima.? Suppose we have a function  $ f:\mathbb{R}^2 \to \mathbb{R}$, of class $C^2$ that satisfies:
$3\frac{\partial^2f}{\partial x^2}(x,y)+4\frac{\partial^2f}{\partial y^2}(x,y)=-1$, for all $(x,y) \in \mathbb{R^2}$ 
How to prove that such $f$ can't have local minima. Is an indirect proof easier?
 A: HINT 
$\frac{\partial^2f}{\partial x^2}<0$
or
$\frac{\partial^2f}{\partial y^2}<0$
what does it tell you? with regards to condition of minimum
if you want more explanation as follows
$\frac{\partial^2f}{\partial x^2} <0$ and $\frac{\partial^2f}{\partial y^2} >0$ then $D<0$
$\frac{\partial^2f}{\partial x^2}>0$ and $\frac{\partial^2f}{\partial y^2}<0$ again $D<0$
if both are negative then its local maxima 
hope that helps
A: At  points $(x,y)$ where $\nabla f(x,y)\ne{\bf 0}$ we cannot have a local minimum. Consider now a point ${\bf p}:=(x_0,y_0)$ with $\nabla f({\bf p})={\bf 0}$.
Since $3f_{xx}({\bf p})+4f_{yy}({\bf p})<0$ by assumption, at least one of $f_{xx}({\bf p})$, $f_{yy}({\bf p})$ is negative. Assume that $c:=f_{xx}({\bf p})<0$ and consider the auxiliary function
$$\phi(t):=f(x_0+t,y_0)\qquad(-h<t<h)\ .$$
Then by Taylor's theorem we have
$$\phi(t)=\phi(0)+\phi'(0)t+{c\over2}t^2+o(t^2)\qquad(t\to0)\ .$$
Since $\phi'(0)=f_x({\bf p})=0$ this implies
$$f(x_0+t,y_0)=f(x_0,y_0)+\left({c\over2}+o(1)\right)t^2\qquad(t\to0)\ ,$$
from which we conclude that $f(x_0+t,y_0)<f(x_0,y_0)$ for all small enough $|t|\ne0$.
