Help understanding power ratings in optocoupler datasheet
The datasheet for the optocoupler is divided in specification for the LED (input) and the phototransistor (output).

You should familiarize yourself with the basic idea of both devices.

Electronics terminology


  I need to know how much voltage and intensity can I support with this optocoupler


There's no term "intensity" in electronics engineering. No one knows what you mean by that. That said, don't confuse \$ I \$ with "intensity", \$ I \$ is current.

Absolute maximum ratings (output)



\$ I_{C} \$ - Maxmimum current the collector withstands regardless of the voltage
\$ V_{CEO} \$ - Maximum voltage between collector and emitter with transistor open (simply put: no light shining on it) before voltage break-though
\$ V_{ECO} \$ - Maximum voltage between emitter and collector (reverse voltage, does not matter in proper operation)
\$ P_{TOT} \$ - Total power dissipation of the whole package (specifies the sum of the dissipated LED and the transistor power)  

Phototransistor power

Under "Absolute Maximum Ratings" the power dissipation (\$ P_{C} \$) is specified as \$ 100mW \$. That is different combinations of collector-emitter voltage times the collector current. That might be \$ 50mA * 2V \$ or \$ 20mA * 5V \$.

The current and the collector-emitter voltage depends on the rest of the circuit. Assuming this test/example circuit:



Simplified: if the voltage \$ V_{cc} \$ were \$ 9V \$ and \$ R_{L} \$ were \$ 1k \$ the maximum current through the corrector would be \$ { 9V \over 1k } = 9mA \$. In practice the collector-emitter junction drops voltage, so the power would be \$ 0.5V * 9mA  = 4.5mW \$ for example.

If your \$ V_{cc} \$ were above \$ 35 V \$, then, no matter how tiny the collector current is (regardless of the collector resistor), you might already damage your phototransistor.

If you were to drive another transistor or a microcontoller input this should suffice. Directly driving a relay for example might already exceed the current rating for the phototransistor. You'd need to drive a power transistor instead then.

Regarding higher drive current see this related question for examples: Optocoupler driving 300mA continuously

LED power


  The input is 6V and 60mA (is it right?). 




Wrong. The specified \$ 6V \$ is the voltage the diode is supposed to sustain when voltage is applied in reverse. See Wikipedia for details on that. If you design your circuit properly this does not apply.

Take a look at the next page. The maximum forward voltage at \$ 20mA \$ current is specified as \$ 1.2V \$. That makes \$ 24mW \$ power dissipation for the LED.

Note that the maximum continuous current can be as high as \$ 60mA \$, as long as you do not drive the LED harder than that you should stay below the \$ 100mW \$ rating. Current and forward voltage are in a direct, exponential relation to each other (you can observe this in the "Forward current vs. Forward voltage" graph in datasheets).



I answered this question since it seems to be well defined, regardless of a possible X/Y situation. It might very well be the case that you don't need an optocoupler at all. You shall search first and possibly ask a specific question on how to achieve X instead.