Have a look at its peak

For rigorous proof : Use following two identity

${ a }^{ 3 }\quad +\quad { b }^{ 3 }\quad =\quad (a\quad +\quad b)({ a }^{ 2 }\quad +\quad { b }^{ 2 }\quad -\quad ab)\\ \\ { a }^{ 2 }\quad +\quad { b }^{ 2 }\quad =\quad { (a\quad +\quad b) }^{ 2 }\quad -\quad 2ab$ .

$f\left( x \right) =(\sin ^{ -1 }{ x } +\cos ^{ -1 }{ x } )({ (\sin ^{ -1 }{ x } ) }^{ 2 }+{ (\cos ^{ -1 }{ x) } }^{ 2 }-\sin ^{ -1 }{ x } \cos ^{ -1 }{ x } )\\ \\ f\left( x \right) =\cfrac { \pi }{ 2 } ({ \cfrac { \pi }{ 4 } }^{ 2 }-3\sin ^{ -1 }{ x } \cos ^{ -1 }{ x } )\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \bullet \quad { (\sin ^{ -1 }{ x } ) }_{ min }\quad =\quad \cfrac { -\pi }{ 2 } \quad \& \quad ({ \cos ^{ -1 }{ x } ) }_{ max }\quad =\quad \pi \quad \quad \quad \\ (at\quad Same\quad x\quad ,\quad i.e\quad x\quad =\quad -1)\\ \\ { f\left( x \right) }_{ max }=f\left( -1 \right) =\cfrac { 7 }{ 8 } { \pi }^{ 3 }\,$ .

Note : Also Here Minimum of f(x) will be

${ f\left( x \right) }_{ min }\quad =\quad f(\cfrac { 1 }{ \sqrt { 2 } } )\quad =\quad \cfrac { { \pi }^{ 3 } }{ 32 } \quad$ .

After some observation, I have found out that if x=-1, f(x) gives the maximum value. And also, it is also notable that for x=-1, arc cosx =pie and it is the maximum value of arc cosx and the value of arc sinx can never be more than pie/2 so this mathematical nature also ensures the maximum value of f(x) when x=-1