Well...

The gradient vector is perpendicular to the level curve!

Proof:
If f[x,y] = c is your level curve and you give it a parametrization r[t]=x[t]i+y[t]j, then g[t]=f[x[t],y[t]]=c.
But then g'[t] = 0
By the chain rule g'[t] = gradf[x[t],y[t]].{x'[t],y'[t]}.
So gradf[x[t],y[t]].{x'[t],y'[t]} = 0.
This means that the gradient is perpendicular to the tangent vector at that point.
By the way, we cannot visualize it here, but this works for any dimension.

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